t-Butyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline [Ind. Eng. Chem. Res., 27, 2224 (1988)]. TBA was produced by the liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The system is normally a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can achieve reasonable miscibility. The reaction mechanism is believed to be
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Complete Question:
t-Butyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline [Ind. Eng. Chem. Res., 27, 2224 (1988)]. TBA was produced by the liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The system is normally a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can achieve reasonable miscibility. The reaction mechanism is believed to be
I + S ⇄ I*S
W + S ⇄ W*S
W*S + I*S ⇄ TBA * S * S
TBA * S ⇄ TBA + S
Derive a rate law assuming:
(a) The surface reaction is rate limiting
(b) The adsorption of isobutene is limiting
(c) The reaction follows Eley-rideal Kinetics
I*S+W ⇄ TBA * S
and surface reaction is limiting
(d) Isobutene (I) and water (W) are adsorbed on different sites
I + S₁ ⇄ I*S₁
W + S₂ ⇄ W*S₂
TBA is not on the surface, and the surface reaction is rate-limiting
[tex][Ans: r'_{TBA}=-r'_1=\frac{k[C_1C_w-C_{TBA/K_C}]}{(1+K_WC_W)(1+K_1C_1)} ][/tex]
(e) What generalizations can you make by comparing rate laws derived from part (a) through (d)?
Answer and explanation:
The mechanism for the production of t-butyl alcohol is as follows:
the reaction and rate law for the adsorption of isobutene over the amberlyst-15 is as follows:
I + S ⇄ I * S [tex]-r_{ADI} = k_I(C_1C_v-\frac{C_{I.S}}{K_I} )[/tex]
where [tex]C_V[/tex] is the concentration of vacant site
[tex]K_I[/tex] is the equilibrium constant of the adsorption
[tex]k_I[/tex] is the rate constant for forward
[tex]C_I,C_{I.S}[/tex] are concentration of isobutene and site filled with isobutene
The reaction and rate law for the adsorption of water (W) over the amberlyst-15 catalyst catalyst is as follows
W + S ⇄ W.S [tex]-r_{ADW} = k_W(C_WC_V-\frac{C_{W.S}}{K_W} )[/tex]
The reaction and rate law for the surface reaction on the catalyst is as follows
W.S + I.S ⇄ TBA . S + Sn [tex]-r_s = k_s(C_{W.S}C_{I.S}-\frac{C_{TBA.S}C_V}{K_s} )[/tex]
The reaction and rate law for the desorption of TBA from catalyst is as follows
TBA . S ⇄ TBA + S [tex]-r_{D TBA} = k_{DTBA}(C_{TBA.S}-\frac{C_{TBA}C_V}{K_{DTBA}} )[/tex]
the attached image below gives the remaining steps
TBA, an octane enhancer, is produced by the hydration of isobutene over an Amberlyst-15 catalyst, resulting in a multiphase mixture that can be made miscible with the help of cosolvents or excess TBA.
Explanation:The production of t-Butyl Alcohol (TBA) involves a liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The question provides background information on the reaction mechanism and mentions the use of cosolvents or excess TBA to achieve reasonable miscibility in the multiphase mixture. This process results in a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can aid in achieving miscibility. Essentially, the production of TBA serves as a safer alternative to lead additives in gasoline, enhancing octane levels.
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Related Questions
1. What values are needed to determine the energy of an electron in a many‑electron atom?
O n
O ????
O m????
O ms
2. What information is most important in determining the size of an orbital?
O n
O ????
O m????
O ms
3. What information is needed to determine the orientation of an orbital?
O n
O ????
O m????
O ms
4. What information is needed to determine the general shape of an orbital?
O n
O ????
O m????
O ms
Answers
Answer: 1. n
2. n
3. [tex]m_s[/tex]
4. l
Explanation:
Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1).For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as [tex]m_s[/tex] . The value of this quantum number ranges from -l to +l.
Spin Quantum number : It describes the direction of electron spin. This is represented as s.
To determine the energy of an electron in a many-electron atom, the principal and angular momentum quantum numbers are needed (n and l). The principal quantum number (n) primarily determines the size of an orbital, while the magnetic quantum number (ml) is needed for its orientation. The angular momentum quantum number (l) is responsible for the general shape of an orbital.
To determine the energy of an electron in a many-electron atom, the following quantum numbers are needed:
n (principal quantum number) determines the general range for the value of energy and the probable distances that the electron can be from the nucleus.I (angular momentum quantum number) helps to determine subshell energy levels.The most important information in determining the size of an orbital is:
n (principal quantum number), which largely determines the energy and size of the orbital.To determine the orientation of an orbital, the quantum number needed is:
ml (magnetic quantum number), which describes the orientation of the orbital in space.The quantum number needed to determine the general shape of an orbital is: I (angular momentum quantum number), which describes the shape or type of the orbital.
Each electron also has a spin quantum number, ms, which determines the spin of an electron and can have a value of either +1/2 or -1/2.
Aldol condensation of 2,5-heptanedione yields a mixture of two enone products in a 9:1 ratio. Treatment of the minor product with aqueous NaOH converts it into the major product; the interconversion proceeds as follows: Hydroxide ion adds to the double bond, forming enolate ion 1; Proton transfer occurs, yielding tetrahedral intermediate 2; Ring opening occurs, yielding enolate ion 3; Protonation of enolate ion 3 occurs, yielding 2,5-heptanedione; Deprotonation at C-6 occurs, yielding enolate ion 5; Enolate ion 5 attacks C-2, yielding tetrahedral intermediate 6; Protonation occurs to yield aldol addition product 7; Dehydration yields the more stable product.
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Complete Question
The complete question is shown on the first uploaded image
Answer:
The diagram of the mechanism of this reaction is shown on the second uploaded image
The structure of the enolate Ion 1 is shown on the third uploaded image
Explanation:
The aldol condensation of 2,5-heptanedione involves several steps of proton transfer and structural changes to yield stable products. These steps showcase the dynamic nature of chemical reactions and examples of sequential proton transfers seen in polyprotic acids.
Explanation:The aldol condensation of 2,5-heptanedione yielding a mixture of enone products is a chain of reactions involving the transfer of protons. Iterating the process you described, it starts with hydroxide ion adding to the double bond to form intermediate enolate ion 1, followed by proton transfer resulting in tetrahedral intermediate 2. The ring opening process then forms another enolate ion (3), which after protonation, yields the initial 2,5-heptanedione. A deprotonation at C-6 subsequently yields enolate ion 5, which attacks C-2 to generate tetrahedral intermediate 6. Protonation at this stage will yield aldol addition product 7. Dehydration of this product will finally yield the more stable product. This sequence is an example of sequential proton transfers often seen in polyprotic acids.
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a gas at 80kPa occupies a volume of 5mL. WHat volume will the gas occupy at 70kPa?
Answers
Answer:
Volume = 5.71mL
Explanation:
Applying Boyle's law
P1V1= P2V2
P1= 80kPa, V1= 5mL,P2= 70kPa, V2=?
Substitute into above formula
80×5= 70×V2
V2= (80×5)/70 = 5.71mL
An atom has the following electron configuration 1s2 2s2 2p6 3s2 3p4 . How many valence electrons does this Atom have
Answers
Answer:
6
Explanation:
This atom is sulfur (if the electrons are equal to the protons/not an ion). You can tell the number of valence electrons by looking at the individual shell. The first shell (1s) can only hold 2 electrons. The second shell (2s and 2p) can hold 8 electrons. The third shell (3s and 3p), which is the valence shell, only has 6 out of its possible 8 electrons, so this atom has 6 valence electrons.
The number of valence electron that the atom with the electronic configuration of 1s²2s²2p⁶3s²3p⁴ has is 6
Valence electron(s) are the electrons located on the outermost shell of an atom.
With the above information in mind, we shall determine the number of valence electron that the atom has. This is illustrated as follow:
Electronic configuration => 1s²2s²2p⁶3s²3p⁴
Valence shell => 3s²3p⁴
Valence electron = 2 + 4
Valence electron = 6Therefore, the atom has 6 valence electrons.
See attachment image
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We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of
sodium chloroacetate (ClCH2COONa) in 1.0 L of water (pKa = 2,865).
to. Calculate the pH
yes. Calculate the pH using the formal forms (activities). Have on
counts the contribution of the protons (section a) in the calculation of the ionic strength.
C. Find the pH of a mixture prepared by dissolving the following compounds
in a final volume of 1L: 0.08 moles of ClCH2COOH, 0.04 moles of
ClCH2COONa, 0.05 moles of HNO3 and 0.06 moles of NaOH
Answers
Answer:
A. pH using molar concentrations = 2.56
B. pH using activities = 2.46
C. pH of mixture = 2.56
Explanation:
A. pH using molar concentrations
ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺
HA + H₂O ⇌ A⁻ + H₃O⁺
We have a solution of 0.08 mol HA and 0.04 mol A⁻
We can use the Henderson-Hasselbalch equation to calculate the pH.
[tex]\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}[/tex]
B. pH using activities
(i) Calculate [H⁺]
pH = -log[H⁺]
[tex]\text{[H$^{+}$]} = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73 \times 10^{-3}\text{ mol/L}[/tex]
(ii) Calculate the ionic strength of the solution
We have a solution of 0.08 mol·L⁻¹ HA, 0.04 mol·L⁻¹ Na⁺, 0.04 mol·L⁻¹ A⁻, and 0.00273 mol·L⁻¹ H⁺.
The formula for ionic strength is
[tex]I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} + 0.00273\times(+1)^{2}\right]\\\\= \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041[/tex]
(iii) Calculate the activity coefficients
[tex]\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79[/tex]
(iv) Calculate the initial activity of A⁻
a = γc = 0.79 × 0.04= 0.032
(v) Calculate the pH
[tex]\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\[/tex]
C. Calculate the pH of the mixture
The mixture initially contains 0.08 mol HA, 0.04 mol Na⁺, 0.04 mol A⁻, 0.05 mol HNO₃, and 0.06 mol NaOH.
The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.
The excess NaOH will react with 0.01 mol HA to form 0.01 mol Na⁺ and 0.01 mol A⁻.
The final solution will contain 0.07 mol HA, 0.10 mol Na⁺, 0.05 mol A⁻, and 0.05 mol NO₃⁻.
(i) Calculate the ionic strength
[tex]I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} + 0.05\times(-1)^{2}\right]\\\\= \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10[/tex]
(ii) Calculate the activity coefficients
[tex]\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.10} = -0.510\times 0.32 = -0.16\\\gamma = 10^{-0.16} = 0.69[/tex]
(iii) Calculate the initial activity of A⁻:
a = γc = 0.69 × 0.05= 0.034
(iv) Calculate the pH
[tex]\text{pH} = 2.865 + \log \left(\dfrac{0.034}{0.07}\right ) = 2.865 + \log 0.49 = 2.865 - 0.31 = \mathbf{2.56}[/tex]
Which statement best describes what is taking place? Copper is being oxidized. Copper is being reduced. Copper is losing electrons. Copper is a reducing agent.
Answers
Answer:
copper is being reduced
Explanation: I got 100% on the quiz good luck this class is hard lol
Answer:
The answer is the second option
Copper is being reduced
Explanation:
Have a good day just took test
A mixture of propane and butane is fed into a furnace where it is mixed with air. The furnace exhaust leaves the furnace at 337�C, 786.0 mmHg and contains only N2, O2, CO2, and H2O. The partial pressure of O2 in the exhaust is 10.38 mmHg and the partial pressure of CO2 in the exhaust is 88.03 mmHg.
1.What is the mole fraction of propane in the fuel stream?
2.What is the mole fraction of water in the exhaust stream?
3.What is the dew point temperature of the exhaust gas?
Answers
Answer:
1. 3.29mol
2. 0.125molH2o/mol
3. 52.5'C
Explanation: The step by step explanation are attached to the answer.
Answer:
Mole fraction of propane = 0.74 mol C₃H₈/mol
Mole fraction of water = 0.29 mol H₂0 /mol
Dew point temperature = 52.5°C
Explanation:
See the attached file for the calculation.
A precipitate forms when mixing solutions of silver nitrate (AgNO,) and sodium chloride (NaCI). Complete the net ionic equation for this reaction by filling in the blanks. Do not include charges on any ions and do not include phase symbols. Provide your answer below (aq)+(aq)6)
Answers
The reaction between solutions of silver nitrate and sodium chloride results in a precipitate due to a double-replacement reaction. The solid precipitate formed is silver chloride (AgCl). The net ionic equation becomes: Ag+ (aq) + Cl- (aq) → AgCl (s).
Explanation:
When you mix solutions of silver nitrate (AgNO3) and sodium chloride (NaCl), a precipitate forms. This is because of a double-replacement reaction that occurs between the two compounds. You start with AgNO3 (silver nitrate) and NaCl (sodium chloride), and end up with NaNO3 (sodium nitrate) and AgCl (silver chloride). The silver chloride precipitates, or becomes solid, leaving the sodium nitrate in solution. This can be shown by the net ionic equation:Ag+ (aq) + Cl- (aq) → AgCl (s). This equation represents the formation of the silver chloride precipitate. The ions Na+ and NO3- do not take part in the reaction and so are not included in the net ionic equation.
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An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the ratio of the ionic radius of the cation to the ionic radius of the anion is 0.840, what is the ionic radius of the anion
Answers
Answer:
the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]
Explanation:
From the diagram shown below :
The anion [tex]Cl^-[/tex] is located at the corners
The cation [tex]Cs^+[/tex] is located at the body center
The Body diagonal length = [tex]\sqrt{3 \ a }[/tex]
∴ [tex]2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a[/tex]
Given that :
[tex]\frac{r^+}{r^-} =0.84[/tex] (i.e the ratio of the ionic radius of the cation to the ionic radius of
the anion )
[tex]0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a[/tex]
Also ; a = 664 pm
Then :
[tex]r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm[/tex]
Therefore, the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]
The ionic radius of the anion [tex]r^-=312.52pm[/tex]
Primitive cubic structure:The anion is placed on the corners and the cation is placed on the frame center.
The Body diagonal length = [tex]\sqrt{3a}[/tex]
[tex]2r^++2r^-=\sqrt{3a} \\\\r^++r^-=\sqrt{3}/2a }[/tex]
Given:
Ratio= 0.840
[tex]\frac{r^+}{r^-}=0.840[/tex]
[tex]0.84r^-+r^-=\sqrt{3}/2a \\\\1.84r^-=3/2a\\\\r^-=\sqrt{3}/2*1.84a[/tex]
Also ; a = 664 pm
Then : [tex]r^-[/tex] =312.52 pm
Therefore, the ionic radius of the anion = 312.52 pm
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2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l) was studied with the following results: Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828 a. Determine the rate law for the reaction. b. Calculate the value of the rate constant with the proper units. c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.
Answers
Explanation:
2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l)
The data is given as;
Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s)
1 0.060 0.030 0.0248
2 0.020 0.030 0.00276
3 0.020 0.090 0.00828
a) Rate law is given as;
Rate = k [ClO2]^x [OH-]^y
From experiments 2 and 3, tripling the concentration of [OH-] also triples the rate of the reaction. This means the reaction is first order with respect to [OH-]
From experiments 1 and 2, when the [ClO2] decreases by a factor of 3, the rate decreases by a factor of 9. This means the reaction is second order with respect to [ClO2]
Rate = k [ClO2]² [OH-]
b. Calculate the value of the rate constant with the proper units.
Taking experiment 1,
0.0248 = k (0.060)²(0.030)
k = 0.0248 / 0.000108
k = 229.63 M-2 s-1
c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.
Rate = 229.63 [ClO2]² [OH-]
Rate = 229.63 (0.100)²(0.050)
Rate = 0.1148 M/s
Final answer:
The rate law is rate = k[ClO2]^2[OH-], with k ≈ 22.2222 M^-2s^-1. The calculated rate with [ClO2] = 0.100 M and [OH-] = 0.050 M is approximately 0.111 M/s.
Explanation:
To determine the rate law for the reaction given, we should look at the changes in concentration and the effect on the initial rate.
Comparing experiment 1 and 2:
When [ClO2] is changed from 0.060 M to 0.020 M (decreased by a factor of 3), the initial rate goes from 0.0248 M/s to 0.00276 M/s (decreased by a factor of ~9).This suggests that the rate is proportional to [ClO2] squared ([ClO2]^2).Comparing experiment 2 and 3:
When [OH-] is increased from 0.030 M to 0.090 M (tripled), the rate increases from 0.00276 M/s to 0.00828 M/s (tripled).This indicates the rate is first order with respect to [OH-].Therefore, the rate law is: rate = k[ClO2]^2[OH-].
Using experiment 1 data to calculate the rate constant (k):
0.0248 M/s = k(0.060^2)(0.030)k = 0.0248 / (0.060^2 * 0.030)k ≈ 22.2222 M-2s-1To calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M:
rate = 22.2222 M-2s-1 * (0.100)^2 * 0.050rate = 0.111 M/sDescribes the three-dimensional arrangement of the atoms in a molecule states that two negatively charged particles (electrons) will always repel one another with equal and opposite forces ensures bonding through shared valence electron pairs is the illustrated definition of the octet rule but not the duet rule
Answers
Answer:
When atoms other than hydrogen form covalent bonds, an octet is accomplished by sharing. The octet rule can be used to explain the number of covalent bonds an atom forms. This number normally equals the number of electrons that atom needs to have a total of eight electrons (an octet) in its outer shell
Explanation:
chemistry, the octet rule explains how atoms of different elements combine to form molecules. ... In a chemical formula, the octet rule strongly governs the number of atoms for each element in a molecule; for example, calcium fluoride is CaF2 because two fluorine atoms and one calcium satisfy the rule.
octet rule: Atoms lose, gain, or share electrons in order to have a full valence shell of eight electrons. Hydrogen is an exception because it can hold a maximum of two electrons in its valence level.
There is another rule, called the duplet rule, that states that some elements can be stable with two electrons in their shell. Hydrogen and helium are special cases that do not follow the octet rule but the duplet rule. ... They are stable in a duplet state instead of an octet state.
When the epoxide 2-vinyloxirane reacts with lithium dibutylcuprate, followed by protonolysis, a compound A is the major product formed. Oxidation of A with PCC yields B, a compound that gives a positive Tollens test and has an intense UV absorption around 215 nm. Treatment of B with Ag2O, followed by catalytic hydrogenation, gives octanoic acid. Identify A and B.
Answers
Answer:
A --- (E)-oct-2-en-1-o1
B ----(E)-oct-2-enal
Explanation:
See the attached file for the structure.
A 60.0 g aluminum block, initially at 55.00 °C, is submerged into an unknown mass of water at 293.15 K in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 25.00 °C. What is the approximate mass of the water? The specific heat of water is 4.18 J/g . °C. The specific heat of aluminum is 0.897 J/g . °C.
Answers
Answer:
The approximate mass of the water is 80kg
Explanation: Heat lost=heat gained
M1c1(Ʃ)=M2c2(Ʃ)
M1 is mass of aluminum
M2 is the mass of water
C1 is specific heat capacity of aluminum
C2 is specific heat capacity of water
Ʃ is change in temperature.
60 x0.897 x(55-25)=M2 x 4.18 x (25-20.15)
1614.6=20.27M2
M2=79.65
M2=80kg
Lithium has two naturally occurring isotopes: lithium−6 and lithium−7. Lithium−6 has a mass of 6.01512 relative to carbon−12 and makes up 7.47 percent of all naturally occurring lithium. Lithium−7 has a mass of 7.016 compared to carbon−12 and makes up the remaining 92.53 percent. According to this information, what is the atomic weight of lithium?
Answers
Answer:
6.941
Explanation:
Step 1:
Representation.
Let Lithium−6 be isotope A
Let Lithium−7 be isotope B
Let the abundance of isotope A (Lithium-6) be A%
Let the abundance of isotope B (Lithium-6) be B%
Step 2:
Data obtained from the question. This includes:
Mass of isotope A (Lithium−6) =
6.01512
Abundance of isotope A (Lithium−6) = A% = 7.47%
Mass of isotope B (Lithium−7) = 7.016
Abundance of isotope B (Lithium−7) = B% = 92.53%
Atomic weight of lithium =?
Step 3:
Determination of the atomic weight of lithium. This is illustrated below:
Atomic weight = [(Mass of AxA%)/100] + [(Mass of BxB%)/100]
Atomic weight = [(6.01512x7.47)/100] + [(7.016x92.53)/100]
Atomic weight = 0.449 + 6.492
Atomic weight of lithium is 6.941
Final answer:
The atomic weight of lithium is approximately 6.9423 u, which is the weighted average of the masses of its isotopes, lithium-6 and lithium-7, with respect to their natural abundances.
Explanation:
To calculate the atomic weight of lithium, we consider the relative abundances and masses of its naturally occurring isotopes, lithium-6 and lithium-7. The atomic weight is the weighted average of the atomic masses of an element's isotopes, based on their natural abundance. Using the provided information:
Lithium-6 has a mass of 6.01512 and an abundance of 7.47%.
Lithium-7 has a mass of 7.016 and an abundance of 92.53%.
The calculation is as follows:
(6.01512 × 0.0747) + (7.016 × 0.9253) = 0.4487 + 6.4936 = 6.9423
Therefore, the atomic weight of lithium is approximately 6.9423 u (atomic mass units).
Calculate the amount of heat needed to convert 15.0 g of liquid water at 87 ºC to steam at 135. ºC.
45,765 J
815 J
33,900 J
35,807 J
Answers
Answer:
The total heat required is 35807 J
Explanation:
Step 1: Data given
Mass of water = 15.0 grams
Initial temperature of water = 87.0 °C
Temperature of steam = 135.0 °C
ΔHfus = 334 J/g
ΔHVap = 2260 J/g
Step 2: Calculate heat required to heat water from 85 to 100 °C
Q = m*c*ΔT
⇒Q = the heat required = TO BE DETERMINED
⇒m = the mass of water = 15.0 grams
⇒c= the specific heat of water = 4.18 J/g°C
⇒ΔT = the change of temperature = T2 - T1 = 100 - 87 = 13°C
Q = 15.0 * 4.18 J/g°C * 13 °C
Q = 815.1 J
Step 3: Calculate heat required to change water at 100 °C to steam
Q = m * ΔHVap
Q = 15.0 grams * 2260 J/g
Q = 33900 J
Step 4: Calculate heat required to heat steam from 100 °C to 135 °C
Q = m*c*ΔT
⇒Q = the heat required = TO BE DETERMINED
⇒m = the mass of water = 15.0 grams
⇒c= the specific heat of steam = 2.09 J/g°C
⇒ΔT = the change of temperature = T2 - T1 = 135 - 100 = 35°C
Q = 15.0 * 2.09 * 35 °C
Q = 1097.25 J
Step 5: Calculate the total heat required
Q = 35807 J
The total heat required is 35807 J
T
It is desired to determine the concentration of arsenic in a lake sediment sample by means of neutron activation analysis. The nuclide captures a neutron to form , which in turn undergoes β decay. The daughter nuclide produces the characteristic γ rays used for the analysis. What is the daughter nuclide?
Answers
Answer:
The daughter nuclide is selenium; 76:34 Se
Explanation:
The complete question is as follows;
is desired to determine the concentration of arsenic in a lake sediment sample by means of neutron activation analysis. The nuclide 75:33 As captures a neutron to form 76:33 As, which in turn undergoes beta decay. The daughter nuclide produces the characteristic gamma rays used for the analysis. What is the daughter nuclide?
solution:
Please check attachment for decay equations and explanations
Which gas is most abundant in Earth’s atmosphere?
oxygen
nitrogen
water vapor
carbon dioxide
Answers
Answer:
Nitrogen
Explanation: nitrogen is around 70% of the air hope this helps god bless
How many grams of NaCl are needed to prepare 50.0 grams of 35.0% of salt solution?
Answers
Answer:
17.5 g
Explanation:
Given data
Mass of solution to be prepared: 50.0 gramsConcentration of the salt solution: 35.0%The concentration by mass of NaCl in the solution is 35.0%, that is, there are 35.0 grams of sodium chloride per 100 grams of solution. We will use this ratio to find the mass of sodium chloride required to prepare 50.0 grams of a 35.0% salt solution.
[tex]50.0gSolution \times \frac{35.0gNaCl}{100gSolution} = 17.5gNaCl[/tex]
To prepare a 50 gram 35% salt solution, you need 17.5 grams of NaCl.
Explanation:To find out how many grams of NaCl are needed to prepare 50.0 grams of a 35.0% salt solution, you use the definition of percent concentration by mass: (mass of solute/mass of solution) x 100%. The mass of the solution is the total mass, which is the sum of the mass of the solute (NaCl in this case) and the solvent (usually water). For a 35% solution, this means that 35 grams of NaCl are in every 100 grams of solution.
However, we want to prepare 50 grams of solution. So, you set up a ratio: (35 g NaCl/100 g solution) = (x g NaCl/50 g solution). Solving this equation, you find that x = 17.5. Therefore, 17.5 grams of NaCl are needed to prepare a 50 gram 35% salt solution.
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5. You are investigating an arson scene and you find a corpse in the rubble, but you suspect that the victim did not die as a result of the fire. Instead, you suspect that the victim was murdered earlier, and that the blaze was started to cover up the murder. How would you go about determining whether the victim died before the fire
Answers
Answer:
See the answer below.
Explanation:
Fire has three major components:
HeatSmokeGases ( in form of CO, CO2 etc)If the victim had died as a result of the fire, he/he would have inhaled smoke and hot gases from the fire. These components would have resulted in traces of burns and soot deposition in the trachea and lungs as well as traces of CO in the blood of the victim.
If the analysis of the victim's corpse does not reflect some of the results above, it can be effectively concluded that the victim has been dead before the fire.
The single most important indicator of death by the fire would be the presence of CO in the blood of the victim's corpse. All others might be to a less significant degrees.
Final answer:
To determine if the victim died before the fire, forensic anthropologists analyze perimortem injuries, signs of decomposition, and toxicology for smoke inhalation indicators.
Explanation:
To determine whether a victim died before a fire, a forensic anthropologist would analyze the remains for signs of perimortem trauma. This includes examining cut marks or injuries sustained around the time of death which could indicate homicide. Disarticulation of joints and the degree of skeletal preservation can also help establish the time since death, negating the possibility that the fire caused the death if the individual was already deceased and decomposing. Moreover, the presence of smoke inhalation in the lungs can suggest if the person was alive during the fire. Toxicology reports may provide evidence of substances such as carbon monoxide or soot. The presence or absence of such substances can indicate whether death occurred before or during the fire.
Consider the general reaction: 2 A + b B → c C and the following average rate data over a specific time period \Delta t: - \frac{ \Delta A}{\Delta t} = 0.0080 mol L-1 s-1 - \frac{ \Delta B}{\Delta t} = 0.0120 mol L-1 s-1 \frac{ \Delta C}{\Delta t} = 0.0160 mol L-1 s-1 Determine what the coefficient c could be in the balanced equation. Select one: 1 5 2 3 4
Answers
Answer:
c = 4
Explanation:
In general, for the reaction
a A + b B ⇒ c C + d D
the rate is given by:
rate = - 1/a ΔA/Δt = - 1/b ΔB/Δt = + 1/c ΔC/Δt = + 1/d ΔD/Δt
this is done so as to express the rate in a standarized way which is the same to all the reactants and products irrespective of their stoichiometric coefficients.
For this question in particular we know the coefficient of A and need to determine the coefficient c.
- 1/2 ΔA/Δt = + 1/c ΔC/Δt
- 1/2 (-0.0080 ) = + 1/c ( 0.0160 mol L⁻¹s⁻¹ )
0.0040 mol L⁻¹s⁻¹ c = 0.0160 mol L⁻¹s⁻¹
∴ c = 0.0160 / 0.0040 = 4
The hydroxyl radical, a fragment of water vapor known as the "atmospheric cleanser", with a chemical formula of OH, has a globally averaged number density of ~1×10^6 molecules cm-3 that is fairly constant with altitude. Determine the OH mixing ratio (in ppt) at the surface, where total pressure P ~1 atm, temperature T ~282K, and also at 10 km altitude, where P ~0.260 atm,
Answers
Answer:
Check the explanation
Explanation:
The hydroxyl radical, •OH, is the hydroxide ion (OH−) when in neutral form of the Hydroxyl radicals they are extremely reactive (easily becoming hydroxy groups) and as a result are short-lived; however, they form a significant part of radical chemistry. The hydroxyl radical composition is also highly reactive towards oxidative reactions.
Kindly check the attached image below to see the step by step explanation to the question above.
Which of the following descriptions about standard addition and internal standards are NOT correct?
a) Internal standard is useful when the matrix in the unknown is complicated.
b) Standard addition could be a single standard addition or multiple standard additions to an unknown solution.
c) Standard addition is useful when the matrix in the unknown is complicated.
d) Internal standard is used when instrument response varies from run to run.
Answers
Final answer:
Internal standards and standard addition are techniques used in analytical chemistry to ensure accuracy. The incorrect statement is about the use of internal standards.
Explanation:
Internal standards and standard addition are both techniques used in analytical chemistry to ensure the accuracy and reliability of quantitative measurements. In internal standard method, a known amount of a compound is added to all samples and standards, which allows for compensation of errors that may occur during sample preparation and analysis.
Standard addition, on the other hand, involves adding known amounts of a standard solution to an unknown solution to determine the concentration of the analyte of interest. It can be a single standard addition or multiple standard additions, depending on the requirements of the analysis.
Based on the given options, the incorrect description is:
d) Internal standard is used when instrument response varies from run to run.
The correct answer is a) Internal standard is useful when the matrix in the unknown is complicated.
Let's analyze each option to understand why option (a) is incorrect:
a) Internal standard is useful when the matrix in the unknown is complicated.
This statement is incorrect because it confuses the roles of internal standards and standard addition.
An internal standard is a substance that is added in a constant amount to all samples, including the calibration standards and the unknowns. It is used to correct for any variations in the analytical procedure, such as changes in instrument response, sample preparation, and matrix effects that affect the analyte's signal.
However, it is not specifically used because the matrix is complicated; rather, it is used to account for variations in the analytical process. The complexity of the matrix is typically addressed by the standard addition method, which involves adding known quantities of the analyte to the sample to overcome matrix effects.
b) Standard addition could be a single standard addition or multiple standard additions to an unknown solution.
This statement is correct. Standard addition can involve adding a known amount of analyte to the sample once (single standard addition) or several times (multiple standard additions) to construct a calibration curve.
This method is used to compensate for matrix effects that might not be accounted for by using external calibration alone.
c) Standard addition is useful when the matrix in the unknown is complicated.
This statement is correct. The standard addition method is particularly useful for samples with complex matrices that can interfere with the analysis.
By adding known amounts of the analyte directly to the sample, the method allows for the determination of the analyte's concentration while accounting for the matrix effects.
d) Internal standard is used when instrument response varies from run to run.
This statement is correct. An internal standard is used to normalize the response of the analyte and to correct for any variations in the analytical procedure, including changes in instrument response over time.
It helps to ensure the accuracy and precision of the analytical results, regardless of the variations that occur between different runs of the analysis.
Therefore, the statement in option (a) is the one that is not correct, as it misrepresents the use of internal standards in the context of complex matrices.
A sample of gas contains 0.1100 mol of N2(g) and 0.3300 mol of H2(g) and occupies a volume of 20.5 L. The following reaction takes place: N2(g) 3 H2(g) 2 NH3(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant. g
Answers
Final answer:
To find the volume of the sample after the given reaction, Avogadro's law is used to show that the volume of gases at constant temperature and pressure is directly proportional to the number of moles. The balanced equation indicates that two times the volume of NH3 gas is produced from one volume of N2 gas. The final volume of the sample is calculated to be 10.25 L after the reaction.
Explanation:
To determine the volume of the sample after the reaction between N2 and H2, we must first understand the stoichiometry of the reaction. The balanced chemical equation for the synthesis of ammonia is:
N2(g) + 3H2(g) → 2NH3(g)
According to Avogadro's law, at constant temperature and pressure, equal volumes of gases contain an equal number of molecules. Thus, in our reaction, one volume of N2 reacts with three volumes of H2 to produce two volumes of NH3. In this case, we begin with 0.1100 mol of N2 and 0.3300 mol of H2. Since H2 is present in excess (needed only 3 times the amount of N2), all of the N2 will be consumed first.
The stoichiometry tells us that 1 mol of N2 reacts with 3 mol of H2 to form 2 mol of NH3. So 0.1100 mol of N2 would produce 0.2200 mol of NH3. Since we are assuming the temperature and pressure are constant, we can use the ratio of moles to determine the change in volume. Initially, there are 0.1100 + 0.3300 = 0.4400 mol of gases and after reaction we have 0.2200 mol of NH3.
The initial volume is 20.5 L. With the moles reducing from 0.4400 to 0.2200, the volume occupied by the gases after the reaction is expected to be half of the initial volume, given the relationship V1/n1 = V2/n2 where V is volume and n is the number of moles. Therefore, the volume of the sample after the reaction is 10.25 L.
Final answer:
By applying Avogadro's law, we determine that all of the initial N2 and H2 reacts to form NH3. Because equal volumes of any gas at the same temperature and pressure have the same number of molecules, we understand that the volume of NH3 produced will be equivalent to the volume of N2 reactant consumed plus three times this volume due to 3H2 reacting. Therefore, the final volume should be less than the initial volume since 2 moles of NH3 will be produced from 4 moles of combined reactants.
Explanation:
Understanding Reaction Volumes and Avogadro's Law:
Based on the balanced chemical equation N2(g) + 3H2(g) → 2NH3(g), we can apply Avogadro's law to determine the change in volume when the reaction occurs at constant temperature and pressure.
At the onset, we have 0.1100 mol of N2(g) and 0.3300 mol of H2(g) occupying a combined volume of 20.5 L. Assuming the reaction goes to completion, we start by identifying the limiting reactant, the reactant that will be completely consumed in the reaction. The stoichiometry of the reaction indicates that for every 1 mol of N2, 3 mol of H2 are required.
For 0.1100 mol of N2, we would need 0.3300 mol of H2 to fully react based on the stoichiometry of 1:3. Since we have exactly 0.3300 mol of H2, it means N2 is the limiting reactant and all of it will be converted into NH3.
Now, according to Avogadro's law, if one mole of N2 gas occupies a certain volume and reacts to form two moles of NH3 at the same conditions of temperature and pressure, the volume of NH3 produced will double that of the N2 because two moles of NH3 contain double the number of molecules compared to one mole of N2. However, since H2 also takes up volume and we have three moles of H2 reacting for every mole of N2, the total volume initially is the volume of N2 plus three times the volume of N2 (which is the volume of H2).
The stoichiometry tells us that 1 mol of N2 reacts with 3 mol of H2 to form 2 mol of NH3. So 0.1100 mol of N2 would produce 0.2200 mol of NH3. Since we are assuming the temperature and pressure are constant, we can use the ratio of moles to determine the change in volume. Initially, there are 0.1100 + 0.3300 = 0.4400 mol of gases and after reaction we have 0.2200 mol of NH3.
The initial volume is 20.5 L. With the moles reducing from 0.4400 to 0.2200, the volume occupied by the gases after the reaction is expected to be half of the initial volume, given the relationship V1/n1 = V2/n2 where V is volume and n is the number of moles. Therefore, the volume of the sample after the reaction is 10.25 L.
The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constant at 701 KK is measured as 2.57 M−1⋅s−1M−1⋅s−1 and that at 895 KK is measured as 567 M−1⋅s−1M−1⋅s−1. The activation energy is 1.5×102 1.5×102 kJ/molkJ/mol. Predict the rate constant at 525 KK .
Answers
Answer : The rate constant at 525 K is, [tex]0.0606M^{-1}s^{-1}[/tex]
Explanation :
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = rate constant at [tex]701K[/tex] = [tex]2.57M^{-1}s^{-1}[/tex]
[tex]K_2[/tex] = rate constant at [tex]525K[/tex] = ?
[tex]Ea[/tex] = activation energy for the reaction = [tex]1.5\times 10^2kJ/mol=1.5\times 10^5J/mol[/tex]
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = 701 K
[tex]T_2[/tex] = final temperature = 525 K
Now put all the given values in this formula, we get:
[tex]\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}][/tex]
[tex]K_2=0.0606M^{-1}s^{-1}[/tex]
Therefore, the rate constant at 525 K is, [tex]0.0606M^{-1}s^{-1}[/tex]
Nitrate salts (NO3-), when heated, can produce nitrites (NO2-) plus oxygen (O2). A sample of potassium nitrate is heated, and the O2 gas produced is collected in a 730 mL flask. The pressure of the gas in the flask is 2.5 atm, and the temperature is recorded to be 329 K. The value of R= 0.0821 atm L/(mol K)
How many moles of O2 gas were produced?
Answers
Answer:
0.0676 moles of oxygen gas is produced
Explanation:
Step 1: Data given
Volume of the flask = 730 mL = 0.730 L
The pressure of the gas in the flask is 2.5 atm
The temperature is recorded to be 329 K
Step 2: The balanced equation
2NO3- → 2NO2- + O2
Step 3: Calculate moles oxygen gas (O2)
p*V =n * R*T
⇒with p = the pressure of oxygen gas = 2.5 atm
⇒with V = the volume of the flask = 0.730 L
⇒with n = the number of moles O2 gas
⇒with R = the gas constant = 0.0821 L*atm/mol*K
⇒with T ⇒ the temperature = 329 K
n = (p*V) / (R*T)
n = (2.5 * 0.730 ) / (0.0821*329)
n = 0.0676 moles
0.0676 moles of oxygen gas is produced
1. Choose the alkyl halide(s) from the following list of C6H13Br isomers that meet each criterion below. 1) 1-bromohexane 2) 3-bromo-3-methylpentane 3) 1-bromo-2,2-dimethylbutane 4) 3-bromo-2-methylpentane 5) 2-bromo-3-methylpentane a) the compound(s) that can exist as enantiomers b) the compound(s) that can exist as diastereomers c) the compound that gives the fastest SN2 reaction with sodium methoxide d) the compound that is least reactive to sodium methoxide in methanol e) the compound(s) that undergo an SN1 reaction to give rearranged products f) the compound that gives the fastest SN1 reaction
Answers
Answer:
See explanation
Explanation:
the compound that gives the fastest SN2 reaction with sodium methoxide- 1-bromohexane
the compound that gives the fastest SN1 reaction- 3-bromo-3-methylpentane
the compound(s) that undergo an SN1 reaction to give rearranged products- 1-bromo-2,2-dimethylbutane
the compound that is least reactive to sodium methoxide in methanol -
3-bromo-3-methylpentane
the compound(s) that can exist as diastereomers - 3-bromo-3-methylpentane
the compound(s) that can exist as enantiomers- 3-bromo-2-methylpentane
Why would you use a solution, such as a cabbage pH indicator, to measure the pH of household items?
a) to see if food went bad
b) to test the safety of water
c) to make sure conditions are safe
d) to use up old cabbage
Answers
Answer:
-to see if food went bad
-to test the safety of water
-to make sure conditions are safe
Explanation:
Answer:
A, B, C
Explanation:
A chemistry student needs 10.0g of ethanolamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of ethanolamine is ·1.02gcm−3. Calculate the volume of ethanolamine the student should pour out.
Answers
To determine the volume of ethanolamine needed, the student should divide the desired mass (10.0g) by the density of ethanolamine (1.02g/cm³), which gives approximately 9.8 cm³. A trusted source, such as the CRC Handbook of Physics and Chemistry was referenced.
Explanation:To calculate the volume of ethanolamine needed for the experiment, we need to use the formula for density, which is mass/volume. By rearranging the formula to solve for volume gives us volume = mass/density. So, the volume of ethanolamine will be 10.0g / 1.02g/cm3. This calculation gives us approximately 9.80 cm3. Therefore, the student needs to pour out about 9.8 cm3 of ethanolamine for the experiment. The student was able to easily determine this by consulting a trusted resource, the CRC Handbook of Physics and Chemistry.
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Final answer:
To find the volume of ethanolamine needed, divide the mass needed (10.0 g) by the density (1.02 g/cm³) to obtain approximately 9.80 cm³.
Explanation:
To calculate the volume of ethanolamine the student should pour out, we can use the density formula, which is density (d) equals mass (m) divided by volume (V), rearranged to solve for volume. Given that the density of ethanolamine is 1.02 g/cm³ and the student needs 10.0 g of ethanolamine, the equation would be:
V = m / d
Substituting the given values, we get:
V = 10.0 g / 1.02 g/cm³
The calculation results in a volume of approximately 9.80 cm³ of ethanolamine the student should pour out.
When 63.4 g of glycine (C2HNO2 are dissolved in 700. g of a certain mystery liquid X, the freezing point of the solution is 7.9 °C lower than the freezing point of pure X. On the other hand, when 63.4 g of iron(III) chloride are dissolved in the same mass of X, the freezing point of the solution is 13.3 °C lower than the freezing point of pure X Calculate the van't Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits. x 10
Answers
Answer:
3.8 is the van't Hoff factor for iron(III) chloride in X.
Explanation:
[tex]\Delta T_f=i\times K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point =
[tex]K_f[/tex] = freezing point constant
m = molality =[tex]\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}[/tex]
i = van't Hoff factor
we have :
Mass of glycine = 63.4 g
Molar mass of glycine = 71 g/mol
Mass of solvent X = 700. g = 0.7 kg
1 g = 0.001 kg
[tex]K_f[/tex] of solvent X= ?
i = 1 (non electrolyte)
Depression in freezing point= [tex]7.9^oC[/tex]
[tex]7.9^oC=1\times K_f \times \frac{63.4 g}{71 g/mol\times 0.7 kg}[/tex]
[tex]K_f=6.19 ^oC/m[/tex]
When iron(III) chloride is dissolved in 0.7 kg of solvent X
Mass of iron(III) chloride = 63.4 g
Molar mass of iron(III) chloride= 162.5 g/mol
Mass of solvent X = 700. g = 0.7 kg
1 g = 0.001 kg
[tex]K_f[/tex] of solvent X= [tex]6.19 ^oC/m[/tex]
i = ?
Depression in freezing point:[tex]13.3^oC[/tex]
[tex]13.3^oC=i\times 6.19^oC\times \frac{63.4 g}{162.5 g/mol\times 0.7 kg}[/tex]
Solving for i:
i = 3.85 ≈ 3.8
3.8 is the van't Hoff factor for iron(III) chloride in X.
Suppose the formation of iodine proceeds by the following mechanism:
step elementary reaction rate constant
1 H2 (g) + ICl (g) → HI (g) + HCl (g) k1
2 HI (g) + ICl (g) → I2 (g) + HCl (g) k2
Suppose also ki « k2. That is, the first step is much slower than the second. Write the balanced chemical equation for the overall chemical reaction. Write the experimentally- observable rate law for the overall chemical reaction.
Answers
Answer:
Overall reaction
H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)
Overall Rate = k1[H2] [ICl]
Explanation:
Overall reaction
H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)
The overall reaction is the sum of the two two reactions shown in the question. After the two reactions are summed up properly, this overall reaction equation his obtained.
Since K1<<K2 it means that step 1 is slower than step 2. Recall that the rate if reaction depends on the slowest step of the reaction. Hence
Overall Rate = k1[H2] [ICl]
The balanced chemical equation for the overall reactions is H2 (g) + 2ICl (g) → I2 (g) + 2HCl (g). The rate of the reaction is determined by the slowest step, in this case, the first one. So, the experimentally-observable rate law for the overall reaction would be Rate = k1[H2][ICl].
Explanation:The overall reaction is obtained by adding up the given elementary reactions. Adding step 1 and 2, we have:
H2 (g) + 2ICl (g) → 2HI (g) + HCl (g) → H2 (g) + I2 (g) + HCl (g)
After cancelling like terms the balanced chemical equation is:
H2 (g) + 2ICl (g) → I2 (g) + 2HCl (g)
Since the first step is much slower than the second, it's the rate-determining step. The experimentally-observable rate law for the overall reaction will depend on the rate-determining step, hence, Rate = k1[H2][ICl]
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