Answers
Answer:
2-methoxy-2-methylpropane
Explanation:
The first step for this reaction is the carbocation formation. In this step, a tertiary carbocation is formed. Also, we will have a good leaving group so bromide will be formed. Then the methanol acts as a nucleophile and attacks the carbocation. Next, a positive charge is generated upon the oxygen, this charge can be removed when the hydrogen leaves the molecule as [tex]H^+[/tex]. (See figure)
Related Questions
One of the many reactions which occur when iron ore is smelted in a blast furnace is given below. For this reaction, the equilibrium constant in terms of partial pressures, Kp, is 0.900 at 873 K and 0.396 at 1273 K. Assume the reaction takes place in a vessel containing only product and reactant molecules and that AH and AS do not change with temperature. FeO(s) + CO(g) = Fe(s) + CO2(g) a) Calculate AH, AG and AS for the reaction at 873 K b) What is the mole fraction of CO2(g) at 873 K? Include only the gas-phase species in this mole fraction.
Answers
Answer:
Explanation: AH, AG and AS for the reaction at 873 K
delta G =-RT ln Kp
G= -8.314 x 873 ln 0.9
= 764.7J/mol
AG = AH-TAS
AH = 764.7 - 8314(0.9)
Kp =pCO/pCo2
Substitute the values at different temp. Solve simultaneously
Since mole fraction =kp/pCo
Mf =1.5
Answer:
you looking beautiful
A reaction at 17.0 degree C evolves 788.mmol of dinitrogen difluoride gas. Calculate the volume of dinitrogen difluoride gas that is collected. You can assume the pressure in the room is exactly 1atm. Be sure your answer has the correct number of significant digits.
Answers
Answer:
V = 18.8 L
Explanation:
This question is solved by using the ideal gas law :
PV = nRT ∴ V= nRT/P
where n= 0.788 mmol = 0.788 mol (1 mmol is a thousandths of a mol)
R = R constant for ideal gases = 0.08206 Latm/Kmol
T = 17.0 ºC = (17.0 + 273) K = 290 K
and V = volume in liters our unknown
Plugging our values and solving for V,
V = 0.788 mol x 0.08206 Latm/Kmol x 290 K / 1atm = 18.8 L
Applying the Ideal Gas Law with the given number of moles, temperature and pressure shows that the volume of dinitrogen difluoride gas collected is approximately 19.0 L.
Explanation:The volume of a gas can be calculated using the Ideal Gas Law if the number of moles, temperature and pressure are known. In your question, we're given that 788mmol of dinitrogen difluoride gas is evolved at a temperature of 17.0 degree C and a pressure of 1 atm. Firstly, convert the temperature to Kelvins by adding 273.15 to the Celsius temperature, giving us 290.15K.
The Ideal Gas Law is expressed as PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L·atm/ K·mol for this situation) and T is temperature. We can rearrange the equation to solve for volume: V=nRT/P. Therefore, substituting in the known values: V = 0.788 mol * 0.0821 L·atm/ K·mol * 290.15K / 1 atm, we find that the volume of dinitrogen difluoride gas collected is approximately 19.0 L, given correct to the number of significant digits.
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QUICK WILL MARK BRAINLIEST!!!!
to calculate molar mass add the ______ mass for each atom in the chemical _____
from the periodic table.
(fill in the blanks)
Answers
Answer:
To calculate the molar mass add the atomic mass of each atom in the chemical formula.
Explanation:
For example Molar mass of NaHCO₃ calculated as
Molar mass is the sum of masses of all atom present in formula.
Atomic mass of sodium = 23 g/mol
Atomic mass of hydrogen = 1.008 g/mol
Atomic mass of carbon = 12 g/mol
Atomic mas of Oxygen = 16 g/mol
Molar mass of NaHCO₃ = 23 + 1.008 + 12 + 16× 3
Molar mass of NaHCO₃ = 23 + 1.008 + 12 + 48
Molar mass of NaHCO₃ = 84.008 g/mol
Arsine, AsH3, is a highly toxic compound used in the electronics industry for the production of semiconductors. Its vapor pressure is 35 Torr at 111.95 C and 253 Torr at 83.6 C. Using these data, calculate (a) the standard enthalpy of vaporization;
Answers
Answer:
-79.8 × 10⁴ J/mol
Explanation:
Arsine, AsH₃, is a highly toxic compound used in the electronics industry for the production of semiconductors. Its vapor pressure is 35 Torr at 111.95 °C and 253 Torr at 83.6 °C.
Then,
P₁ = 35 torr
T₁ = 111.95 + 273.15 = 385.10 K
P₂ = 253 torr
T₂ = 83.6 + 273.15 = 356.8 K
We can calculate the standard enthalpy of vaporization (ΔH°vap) using the two-point Clausius-Clapeyron equation.
[tex]ln(\frac{P_{2}}{P_{1}} )=\frac{-\Delta H\°_{vap}}{R} .(\frac{1}{T_{2}} -\frac{1}{T_{1}} )[/tex]
where,
R is the ideal gas constant
[tex]ln(\frac{253torr}{35torr})=\frac{-\Delta H\°_{vap}}{8.314J/K.mol} .(\frac{1}{356.8K}-\frac{1}{385.10K})\\ \Delta H\°_{vap}=-79.8 \times 10^{4} J/mol[/tex]
Methane, CH4, reacts with chlorine, Cl2, to produce a series of chlorinated hydrocarbons: methy chloride (CH3Cl), methylene chloride (CH2Cl2), chloroform (CHCl3), and carbon tetrachloride (CCl4). Which compound has the lowest vapor pressure at room temperature? Explain.
Answers
Answer:
CCl₄, because it is the heaviest compound.
Explanation:
When a liquid is in a closed container, evaporation occurs, in what the molecules with the highest kinetic energy can scape of the liquid. The vapor that was formed does pressure on the liquid that remains, and when both phases stay in equilibrium the pressure is called vapor pressure.
We can notice that the vapor pressure is a measure of the volatility of a liquid. Substances with higher vapor pressure are more volatile. The volatility, however, depends on the nature of the forces in the compound and on the molar mass of it.
For the substances given, they are all covalent compounds and have dipole-induced-dipole-induced bonds between the molecules (because they are nonpolar). So, the lowest vapor pressure is in the heaviest compound, which is the most substituted: CCl₄.
Final answer:
Carbon tetrachloride (CCl4) has the lowest vapor pressure at room temperature due to its highest molecular weight and strongest intermolecular forces among the products of the reaction between methane and chlorine.
Explanation:
Methane, CH4, reacts with chlorine, Cl2, to produce chlorinated hydrocarbons such as methy chloride (CH3Cl), methylene chloride (CH2Cl2), chloroform (CHCl3), and carbon tetrachloride (CCl4). Looking at the structures of these molecules, we can see that carbon tetrachloride (CCl4) has the highest number of chlorine atoms, which are highly electronegative and contribute to more significant intermolecular forces like London dispersion forces. Consequently, the compound with the highest molecular weight and strongest intermolecular forces will exhibit the lowest vapor pressure at room temperature, which, in this case, is carbon tetrachloride (CCl4).
For which of the following equations is the change in enthalpy at 25 Celcius and 1 atm pressure equal to ΔHf of CH3OH (l)?
A.) CH3OH + 3/2 O2 ---> CO2 + 2H2O (l)
B.) CH3OH + 3/2 O2 ---> CO2 + 2H2O (g)
C.) 2CH3OH + 3O2 ---> 2CO2 + 4H2O (l)
D.) C + 2H2 + 1/2 O2 ---> CH3OH
Answers
Answer:
The correct option is: D) C (graphite)+ 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l)
Explanation:
The standard enthalpy change of a given chemical reaction ([tex]\Delta H^{\circ }_{r}[/tex]) is equal to the difference of the sum of formation enthalpy ([tex]\Delta H^{\circ }_{f}[/tex]) of products and the sum of formation enthalpy of reactants.
[tex]\Delta H^{\circ }_{r} = \sum v. \Delta H^{\circ }_{f}(products) - \sum v. \Delta H^{\circ }_{f}(reactants)[/tex]
The standard enthalpy of formation is zero for elements that exist in their standard states.
The standard enthalpy change of a reaction is equal to the standard enthalpy of formation of methanol (l) [[tex]\Delta H^{\circ }_{r} = \Delta H^{\circ }_{f}(CH_{3}OH, l)[/tex]], only if:
1. The standard enthalpy of all the other reactants and products should be zero, i.e. all the other reactants and products should be in their standard states.
2. The stoichiometric coefficient of methanol (CH₃OH, l) should be 1.
Among the given options, only the option D satisfies both the conditions.
D) C (graphite)+ 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l)
The standard enthalpy change for this chemical reaction:
[tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ] - \left [\Delta H^{\circ }_{f}(C, s, graphite) + \Delta H^{\circ }_{f}(H_{2}, g) + \Delta H^{\circ }_{f}(O_{2}, g)\right ][/tex]
The standard enthalpy of formation:
[tex]\Delta H^{\circ }_{f}(C, s, graphite) = 0 kJ/mol[/tex]
[tex]\Delta H^{\circ }_{f}(H_{2}, g) = 0 kJ/mol[/tex]
[tex]\Delta H^{\circ }_{f}(O_{2}, g) = 0 kJ/mol[/tex]
⇒ [tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ] - \left [0 kJ/mol + 0 kJ/mol + 0 kJ/mol\right ] [/tex]
⇒ [tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ][/tex]
Therefore, the correct option is (D).
Do sample problem 13.8 in the 8th ed Silberberg book. You add 1.4 kg of ethylene glycol (C2H6O2) antifreeze to 4,192 g of water in your car's radiator. What is the boiling point of the solution? The Kb for water is 0.512 °C/m. Enter to 2 decimal places.
Answers
Answer:
The solution boils at 102.76 °C
Explanation:
Δt = m* Kb x i
i = 1 with 1 particle in solution
m = molality
Kb = 0.512 °C/molal
Step 2: Calculate moles C2H6O2
molar mass of C2H6O2 = 62.068 g/mol
Calculate number of moles:
Moles = Mass / molar mass
Moles = 1400 grams / 62.068 g/mol
Moles C2H6O2 = 22.56 moles
Step 3: Calculate molality
these are in 4192 g of water:
22.56 moles / 4.192 kg water
⇒ moles / kg of water = 5.38 moles / Kg = m olal
Step 4: Calculate ΔT
ΔT= Kb * molal = 5.38 molal* 0.512 °C/m
ΔT = 2.76 °C
Boiling point = 100°C + 2.75 °C = 102.76 °C
the solution boils at 102.76 °C
The boiling point of the ethylene glycol solution is calculated by first finding the molality of the solution and then using this to find the boiling point elevation. The normal boiling point of water is thus raised by this amount, giving a final boiling point of the solution as 102.75 degrees Celsius.
Explanation:To solve this problem, we need to calculate the molality of the ethylene glycol solution and then use this to determine the boiling point elevation of the solution.
Firstly, we need to convert the mass of ethylene glycol (C2H6O2) and water into moles. The molar mass of ethylene glycol is approximately 62.07 g/mol, so 1.4 kg (or 1400 g) of ethylene glycol is equivalent to approximately 22.54 moles. The mass of water is 4192 g, or roughly 4.192 kg.
The molality (m), is thus given by the formula m = moles of solute / kg of solvent, therefore the molality of the ethylene glycol solution is 22.54 moles / 4.192 kg = 5.38 m.
Next, we use the formula for boiling point elevation: ΔTb = Kb * m, where Kb is the ebullioscopic constant for water (given as 0.512 °C/m), and m is the molality calculated earlier. Therefore, the boiling point of the solution rises by ΔTb = 0.512 × 5.38 = 2.75 °C.
The normal boiling point of water is 100 °C, thus, the boiling point of the ethylene glycol solution is 100 °C + 2.75 °C = 102.75 °C.
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A mixture of argon and krypton gases, in a 5.38 L flask at 67 °C, contains 6.18 grams of argon and 9.66 grams of krypton. The partial pressure of krypton in the flask is________atm and the total pressure in the flask is______atm?
Answers
Answer:
The partial pressure of krypton in the flask is 0.59 atm and the total pressure in the flask is 1.39 atm
Explanation:
This must be solved with the Ideal Gas Law equation.
First of all we need the moles or Ar and Kr in the mixture
Moles = Mass / Molar mass
Molar mass Ar 39.95g/m
Moles Ar = 6.18 g/39.95 g/m → 0.154 moles
Molar mass Kr 83.8 g/m
Moles Kr = 9.66 g/ 83.8g/m → 0.115 moles
Total moles in the mixture: 0.154 moles + 0.115 moles = 0.269moles
Now, we have the total moles, we can calculate the total pressure.
P . V = n . R . T
(T° in K = T° in C + 273)
P. 5.38L = 0.269mol . 0.082 L.atm/mol.K . 340K
P = (0.269mol . 0.082 L.atm/mol.K . 340K) / 5.38 L
P = 1.39 atm
Now we have the total pressure, we can apply molar fraction so we can know the partial pressure of Kr.
Kr pressure / Total Pressure = Kr moles / Total moles
Kr pressure / 1.39 atm = 0.115 moles / 0.269 moles
Kr pressure = (0.115 moles / 0.269 moles) / 1.39atm
Kr pressure = 0.59 atm
Consider a chromium-silver voltaic cell that is constructed such that one half-cell consists of the chromium, Cr, electrode immersed in a Cr(NO3)3 solution, and the other half-cell consists of the silver, Ag, electrode immersed in a AgNO3 solution. The two electrodes are connected by a copper wire. The Cr electrode acts as the anode, and the Ag electrode acts as the cathode. To maintain electric neutrality, you add a KNO3 salt bridge separating the two half-cells. Use this information to answer the following The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell is the site of the reduction reaction. Type the half-cell reaction that takes place at the anode for the chromium-silver voltaic cell. Indicate the physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not forget to add electrons in your reaction.
Answers
Answer:
Anode: Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Explanation:
In the anode takes place the oxidation, in which the reducing agent loses electrons.
In the cathode takes place the reduction, in which the oxidizing agent gains electrons.
The corresponding half-reactions are:
Anode: Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
Mutations in genes encoding cell signaling proteins contribute to many cancers. For example, a chromosomal translocation fuses the Bcr gene to the Abl gene, leading to a constitutively active Bcr–Abl kinase and chronic myeloid leukemia (CML). CML is successfully treated with the drug imatinib (trade name Gleevec), which mimics an Abl kinase substrate and thus inhibits kinase activity. What normal cell molecule, and kinase substrate, does Gleevec mimic?
Answers
Imatinib is a small molecule kinase inhibitor. The BCR-ABL kinase can phosphorylate a series of downstream substrates, leading to proliferation of mature granulocytes. Bcr-Abl kinase substrate is the tyrosine. The Protein Tyrosine Kinase activity is an important requirement for malignant transformation, and that it cannot be complemented by any downstream effector, though not all interactions of BCR-ABL with other proteins are phosphotyrosine dependent.
Calculate the hydronium ion, [ H30+], and hydroxide ion, [OH-], concentrations for a
0.0117 M HCl Solution.
Answers
Answer:
[H3O+] = 0.0117 M
[OH-] = 8.5 * 10^-13 M
Explanation:
Step 1: Data given
Concentration of HCl = 0.0117 M
Step 2:
HCl is a strong acid
pH of a strong acid = -log[H+] = - log[H3O+]
[H3O+] = 0.0117 M
pH = -log(0.0117)
pH = 1.93
pOH =14 - 1.93 = 12.07
pOH = -log[OH+] = 12.07
[OH-] = 10^-12.07 = 8.5 * 10^-13
Or
Kw / [H3O+] = [ OH-]
10^-14 / 0.0117 = 8.5*10^-13
Final answer:
The hydronium ion concentration [H3O+] in a 0.0117 M HCl solution is 0.0117 M, and the hydroxide ion concentration [OH-] is calculated to be 8.55 × 10^-13 M using the water dissociation constant.
Explanation:
To calculate the hydronium ion concentration, [H3O+], of a 0.0117 M HCl solution, we first need to understand that HCl is a strong acid which dissociates completely in water. This means that for every mole of HCl dissolved, there will be one mole of H3O+ ions in solution. Hence, the hydronium ion concentration in a 0.0117 M HCl solution is also 0.0117 M.
As for the hydroxide ion concentration, [OH-], we use the water dissociation constant (Kw), which is 1.0 × 10-14 at 25 °C. The product of the concentrations of the hydronium and hydroxide ions in any aqueous solution is equal to Kw. We can find [OH-] by rearranging the expression Kw = [H3O+][OH-] to solve for [OH-], giving us [OH-] = Kw / [H3O+]. Substituting in the values we get [OH-] = 1.0 × 10-14 M / 0.0117 M = 8.55 × 10-13 M.
Assuming 1 mol of Fe3+ and 2 mol of SCN- were allowed to react and reach equilibrium. 0.5 mol of product was formed. The total volume at equilibrium was 1 L. How much Fe3+ remained at equilibrium? (in mol) (1 point) Your Answer: How much SCN- remained at equilibrium? (in mol) (1 point) Your Answer: What’s the equilibrium constant, Kc? (1 point)
Answers
Answer:
a. 0.5 mol
b. 1.5 mol
c. 0.67
Explanation:
Fe3+ + SCN- -----> [FeSCN]2+
a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium
b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium
c. KC = 0.5/(0.5*1.5) = 0.67
In the laboratory a student uses a "coffee cup" calorimeter to determine the specific heat of a metal. She heats 19.6 grams of zinc to 98.37°C and then drops it into a cup containing 82.9 grams of water at 24.16°C. She measures the final temperature to be 25.70°C. Assuming that all of the heat is transferred to the water, she calculates the specific heat of zinc to be __ J/g°C. g
Answers
Answer:
The specific heat of zinc is 0.375 J/g°C
Explanation:
Step 1: Data given
Mass of zinc = 19.6 grams
Mass of water = 82.9 grams
Initial temperature of zinc T1= 98.37 °C
Initial temperature of water T1= 24.16 °C
Final temperature of water (and zinc) T2 = 25.70 °C
Specific heat of water = 4.184 J/g°C
Step 2: Calculate Specific heat of zinc
Q=m*c*ΔT
Qzinc = -Qwater
m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)
⇒ with mass of water = 82.9 grams
⇒ with C(water) = 4.184 J/g°C
⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54
⇒ with mass of zinc = 19.6 grams
⇒ with C(zinc) = TO BE DETERMINED
⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C
Qzinc = -Qwater
m(zinc)*c(zinc)* ΔT(zinc) = - m(water)*c(water)* ΔT(water)
19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C
-1424.332*C(zinc) = -534.155
C(zinc) = 0.375 J/g°C
The specific heat of zinc is 0.375 J/g°C
Consider the following equilibrium.2 NOBr(g)<=> 2 NO(g) + Br2(g)If nitrosyl bromide, NOBr, is 34 percent dissociated at 25°C and the total pressure is 0.25 atm, calculate Kp and Kc for the dissociation at this temperature.
Answers
Answer:
[tex]K_{p}[/tex] of the reaction is [tex]9.521 \times 10^{-3}[/tex].
[tex] K_{c}[/tex] of the reaction is [tex]3.89 \times 10^{-4}[/tex].
Explanation:
Initial pressure of NOBr is "x" atm.
34% = 0.34
The equilibrium chemical reaction is a follows.
[tex]2NOBr(g)\leftrightarrow 2NO(g)+Br_{2}(g)..............(1)[/tex]
Initial x 0 0
Change -0.34x +0.34x +0.17x
Final (x-0.34x) +0.34x +0.17x
Total pressure = sum of the pressure at equilibrium.
[tex]0.25atm=x-0.34x+0.34x+0.17x[/tex]
[tex]0.25atm= 1.17x[/tex]
[tex]x= \frac{0.25}{1.17}=0.213[/tex]
Partial pressure of NOBr= x-0.34x =x (1-0.34)
= 0.213(0.66) = 0.14 atm.
Hence, partial pressure of [tex][]P_{NOBr}][/tex] is 0.14 atm.
[tex]P_{NO}=0.34x = 0.34 \times 0.213 = 0.072[/tex]
[tex]P_{Br_{2}}=0.17x = 0.17 \times 0.213 = 0.036\,atm[/tex]
From equation (1)
[tex]K_{p}= \frac {P^{2}_{NO} P_{Br_{2}}}{P^{2}_{NOBr}}[/tex]
[tex]= \frac{(0.072)^{2} (0.036)}{(0.14)^{2}}= 9.521 \times 10^{-3}[/tex]
Therefore, [tex]K_{p}[/tex] of the reaction is [tex]9.521 \times 10^{-3}[/tex].
[tex]K_{p}= K_{c}(RT)^{\Delta n}[/tex]
Rearrange the equation is as follows.
[tex]K_{c}= \frac{K_{p}}{(RT)^{\Delta n}}[/tex]
[tex]\Delta n[/tex] = number of moles of reactants - Number of moles of products
= 3-2 = 1
[tex]= \frac{9.521 \times 10^{-3}}{(0.0821 \times 298)^{1}}= 0.389 \times 10^{-3} = 3.89\times 10^{-4}[/tex]
Therefore,[tex] K_{c}[/tex] of the reaction is [tex]3.89 \times 10^{-4}[/tex].
The [tex]K_p[/tex] and [tex]K_c[/tex] is " [tex]0.009649615 \ atm[/tex] and [tex]0.00039\ mol\ L^{-1}[/tex]"for the dissociation at this temperature.
Equation:[tex]2NOBr\rightleftharpoons 2NO+Br_2[/tex]
When the initial pressure of [tex]NOBr[/tex] is p, then at equilibrium, therefore the dissociation percentage of [tex]NOBr \ \ is\ \ 34\%[/tex]
[tex]\to P(NOBr)=0.66\ p\\\\\to P(NO)=0.34\ p\\\\\to P(Br_2)= \frac{0.34p}{2}=0.17\ p\\\\[/tex]
Calculating the total pressure:
[tex]\to 0.66\ p+0.34\ p+0.17\ p=1.17\ p=0.25\ atm\\\\[/tex]
[tex]\to p=0.214 \ atm\\\\[/tex]
The equilibrium partial pressures:
[tex]\to P(NOBr)=0.66\ p=0.141\ atm\\\\\to P(NO)=0.34\ p=0.073\ atm\\\\\to P(Br_2)=0.17\ p=0.036\ atm\\\\[/tex]
Calculating the [tex]K_p[/tex]:
[tex]\to K_p =\frac{(P_{NO})^2 (P_{Br_2})}{(P_{NOBr})^2}\\\\[/tex]
[tex]=\frac{(0.073)^2 \times (0.036)}{(0.141)^2}\\\\=\frac{(0.005329) \times (0.036)}{(0.019881)}\\\\=\frac{(0.000191844)}{(0.019881)}\\\\=0.009649615 \ atm[/tex]
Calculating the [tex]K_c[/tex]:
Using formula:
[tex]\to \Delta ng = \text{Number of moles of product - Number of moles of reactant}[/tex]
[tex]= (2+1) - 2\\\\= 3 - 2\\\\=1 \\[/tex]
The relationship between [tex]K_p[/tex] and [tex]K_c[/tex] is:
[tex]\to K_p = K_C(RT)^{\Delta ng}[/tex]
Hence, by putting the values of [tex]K_p[/tex], gas constant [tex]R, T,[/tex] and [tex]\Delta ng[/tex], we can calculate the value of [tex]K_c[/tex].
[tex]\to 0.0096\ atm = - K_c(0.0821 \ atm \ mol L^{-1}K^{-1} \times 298 K)^1[/tex]
[tex]\to K_c = \frac{0.0096 }{0.0821\times 298}[/tex]
[tex]= \frac{0.0096}{ 24.46}\\\\ = 0.00039\ mol\ L^{-1}[/tex]
Therefore, the [tex]K_p[/tex] and [tex]K_c[/tex] is " [tex]0.009649615 \ atm[/tex] and [tex]0.00039\ mol\ L^{-1}[/tex]"for the dissociation at this temperature.
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1-chloro-3-methyl-2-pentene undergoes hydrolysis in warm water to give a mixture of 3-methyl-2-penten-1-ol and 3-methyl-1-penten-3-ol. Draw the structure of the intermediate's resonance contributor leading to the formation of 3-methyl-2-penten-1-ol.
Answers
Answer:
See the figure
Explanation:
As the first step, the Cl leaves the molecule and the double bond is moved to obtain a tertiary carbocation (the most stable one). Then the molecule of water attacks the primary carbon of the double bond, therefore the double bond moves again to the initial place. Finally, a hydronium ion is produced to remove the positive charge of the oxygen.
Which of the following is NOT a proper IUPAC name?
A. 3-pentanone
B. 1-cyclopentanone
C. 2-pentanone
D. 1-pentanone
Answers
Answer:
The IUPAC of 1-pentanone is not correct.
Explanation:
The structures of each of the given compounds with correct IUPAC names are shown in the figure.
In case of ketones there must be two alkyl groups attached to the carbonyl carbon. In case of aldehydes there must be atleast one hydrogen attached to carbonyl carbon.
Now if we are saying that 1-pentanone, it means we are actually naming the compound with wrong functional group. This is infact an aldehyde with the correct name as "pentanal".
Rest of the given IUPAC names are correct.
When heated, lithium reacts with nitrogen to form lithium nitride: 6Li(s) + N2(g) → 2Li3N(s) What is the theoretical yield of Li3N in grams when 12.7 g of Li is heated with 34.7 g of N2? g If the actual yield of Li3N is 5.85 g, what is the percent yield of the reaction? %
Answers
The theoretical yield of Li3N is 1.24 mol or 138.16 g. The percent yield of the reaction is 4.23%.
Explanation:To find the theoretical yield of Li3N, we first need to determine the limiting reactant. This is done by comparing the moles of each reactant to the stoichiometric ratio of the balanced equation. The molar mass of Li is 6.94 g/mol and the molar mass of N2 is 28.02 g/mol. First, convert the given masses of Li and N2 to moles using their respective molar masses:
12.7 g Li X (1 mol Li / 6.94 g Li) = 1.83 mol Li
34.7 g N2 X (1 mol N2 / 28.02 g N2) = 1.24 mol N2
Then, divide the number of moles of each reactant by its stoichiometric coefficient to find the mole ratio:
1.83 mol Li / 6 = 0.305 mol Li3N
1.24 mol N2 / 1 = 1.24 mol Li3N
Since N2 gives a smaller value, it is the limiting reactant. Therefore, the theoretical yield of Li3N is 1.24 mol or 138.16 g.
To calculate the percent yield, divide the actual yield by the theoretical yield, then multiply by 100:
Percent yield = (Actual yield / Theoretical yield) X 100
Percent yield = (5.85 g / 138.16 g) X 100 = 4.23%
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Predict the product of the following reaction: CH3CH=CHCH3+H2OH3PO4⟶product. Enter the IUPAC name of the product
Answers
The product of the given reaction, an acid-catalyzed hydration of an alkene, would be 2-butanol, as per Markovnikov's Rule. This rule predicts the placement of the hydrogen and halide groups in a reaction.
Explanation:The reaction given is an example of an acid-catalyzed hydration of an alkene. In such reactions, an alkene reacts with water in the presence of an acid (in this case, H3PO4) to form an alcohol. The prediction of the product involves recognizing the reaction type and the reagents involved.
For the given reaction: CH3CH=CHCH3 + H2OH3PO4⟶, the product would be 2-butanol.
But why does this process form 2-butanol? It all comes down to Markovnikov's rule, which predicts that in the addition of a protic acid HX to an alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents.
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Saturated liquid water flows steadily into a well-insulated electrical water heater (see Anim. 4-1-1) with a mass flow rate (m⋅) of 1 kg/s at 100 kPa. Determine (a) the electrical power consumption (W⋅el), and (b) the rate of entropy generation (S⋅gen) in the water heater's universe if the heater turns water into saturated vapor at the exit. Assume no pressure loss, neglect changes in ke and pe, and use the PC model. The ambient atmospheric conditions are 100 kPa and 20oC.
Answers
a. 2257.7 kW b. 6.057kw/K
Steam existing at 100KPa with t = 20C, we can obtain values of enthalpy and entropy from table
h1 = hf = 417.4 kJ/kg
s1 = 1.302 KJ/kg.K
h2 = hg = 2675.1 kJ/kg
s2 = 7.359 kJ/K
a. Electrical Power consumption of system is given by Wel = mass x Change in Enthalpy
Wel = 1 x (h2 - h1) = 1 x (2675.1 - 417.4) = 2257.7 kw
b . Entropy Generation is given by Change in Entropy = Entropy generation - heat dissipated/temperature, since we have a well insulated system with no losses, q =0, hence
ΔS = Sgen + 0/T = Sgen
m(S2 - S1) = Sgen
Sgen = 1 x (7.359 - 1.302) = 6.057 kW/K
When Fe2O3(s) reacts with H2(g) according to the following reaction, 2.00 kJ of energy are evolved for each mole of Fe2O3(s) that reacts. Complete the following thermochemical equation. 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g)
Answers
Answer:
3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g) ΔH° = -6.00 kJ
Explanation:
Let's consider the following balanced equation.
3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g)
When 1 mole of Fe₂O₃(s) reacts, 2.00 kJ of energy are evolved. Energy is an extensive property. In the balanced equation there are 3 moles of Fe₂O₃(s), so the evolved energy is:
[tex]3molFe_{2}O_{3}.\frac{2.00kJ}{1molFe_{2}O_{3}} =6.00kJ[/tex]
By convention, when energy is evolved it takes the negative sign. At constant pressure, the thermochemical equation is:
3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g) ΔH° = -6.00 kJ
where
ΔH° is the standard enthalpy of reaction (heat released at constant pressure)
Which of the following represents a rule for balancing a nuclear equation?
I. The total number of protons plus neutrons in the products and reactants must be the same
II. The total number of each type of element in the products and reactants must be the same
III. The total number of nuclear charges in the products and reactants must be the same
IV. The total number of each type of elementary particle in the products and reactants must be the same.
Answers
Answer:
The correct answer is 1 the total number of proton plus neutrons in the products and reactants must be same.
Explanation:
During a radioactive reaction the nucleus of the radioactive atom undergo disintegration or breakdown to form new nucleus.
Radioactive disintegration deals with the emission of α particle or β particle from a radioactive nucleus resulting in the formation of new nucleus.Although new nucleus is formed the total number of proton plus neutrons in product and reactant must be same.
A rule for balancing a nuclear equation is: I. total number of protons + neutrons in the products and reactants must be the same.
What is a Nuclear Equation?A nuclear equation for a radioactive reaction, shows the reactants and product, where the proton number and atomic mass are conserved, unlike in chemical equations.
In a radioactive decay, there is emission of α particle or β particle from a radioactive nucleus to form a new nucleus.
Therefore, a rule for balancing a nuclear equation is: I. total number of protons + neutrons in the products and reactants must be the same.
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When 4.51 g of CaCl2 dissolved in 50.00 mL of water in a coffee cup calorimeter, the temperature of the solution rose from 22.6°C to 25.8°C.
Specific heat of the solution is equal to the specific heat of water = 4.18 J/gºC.
Density of the solution is equal to the density of water = 1.00 g/mL.
What is qsolution?
What is qreaction ?
What is ÎHrxn in kJ/mol of CaCl2 ?
Answers
Explanation:
The heat gained by the solution = q
[tex]q=mc\times (T_{final}-T_{initial})[/tex]
where,
q = heat gained = ?
c = specific heat of solution= [tex]4.18 J/^oC[/tex]
Mass of the solution(m) = mass of water + mass of calcium chloride
Mass of water = ?
Volume of water = 50.00 mL
Density of water = 1.00 g/mL
Mass = Density × Volume
m = 1.00 g/mL × 50.00 mL = 50.00 g
Mass of the solution (m) = 50.00 g + 4.51 g =54.51 g
[tex]T_{final}[/tex] = final temperature = [tex]25.8 ^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]22.6 ^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=54.51 g\times 4.18 J/g^oC\times (25.8-22.6)^oC[/tex]
[tex]q=729.126 J[/tex]
The heat gained by the solution is 729.126 J.
Heat energy released during the reaction = q'
q' = -q ( law of conservation of energy)
q' = -729.126 J
The heat energy released during the reaction is -729.126 J.
Moles of calcium chloride, n = [tex]\frac{4.51 g}{111 g/mol}=0.04063 mol[/tex]
[tex]\Delta H_{rxn}=\frac{q'}{n}=\frac{-729.126 J}{0.04063 mol}=-17,945.23 J/mol= -17.945 kJ/mol[/tex]
The ΔH of the reaction is -17.945 kJ/mol.
Which solution has the highest pH?
A) 0.10 M HBr(aq)
B) 0.10 M HI(aq)
C) 0.10 M HF(aq)
D) 0.10 M HCl(aq)
E) 0.10 M HClO4(aq)
Answers
All solutions A through E, being strong acids at the same concentration, will have virtually the same pH value under the same conditions. None will have a significantly higher pH. In general, the solution with the highest pH would be the one derived from the weakest acid or strongest base.
Explanation:The pH scale is used to determine the acidity or basicity of a solution. The solutions given are all acidic, being solutions of hydrobromic acid (HBr), hydroiodic acid (HI), hydrofluoric acid (HF), hydrochloric acid (HCl), and perchloric acid (HClO4), respectively. Each of these acids dissociates in water to a different extent, affecting the pH of the solution. However, common to all of these acids is that they are strong acids and will fully dissociate in water.
In this case, all solutions have the same molarity (concentration) of 0.10 M. Because all the acids are strong acids and will fully dissociate, the pH of these solutions will be virtually the same. Therefore, none of the solutions will have a significantly higher pH than the others if measured under the same conditions.
In general, to identify the solution with the highest pH (or lowest acidity), you would look for the weakest acid or the strongest base out of the options provided.
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Among the given solutions, C) 0.10 M HF(aq) has the highest pH because HF is a weak acid and only partially dissociates in water, resulting in a higher pH compared to the strong acids HCl, HI, HBr, and HClO₄.
To determine which solution has the highest pH, we need to understand that pH is a measure of the hydrogen ion concentration in a solution. A lower pH signifies a higher hydrogen ion concentration. Since all the given solutions are acids, we must determine which one produces the least amount of hydrogen ions (H⁺).
HCl, HI, HBr, and HClO₄ are strong acids, meaning they fully dissociate in water. This full dissociation results in a high concentration of H⁺ ions, giving them all a pH of 1 for a 0.10 M solution:
HCl: pH = 1.0
HI: pH = 1.0
HBr: pH = 1.0
HClO4: pH = 1.0
HF, however, is a weak acid and only partially dissociates in water. This partial dissociation results in fewer H+ ions compared to the strong acids listed. Consequently, the pH of a 0.10 M HF solution will be higher (less acidic) than the pH of the other solutions.
Given this information, the solution that has the highest pH is:
C) 0.10 M HF(aq)
Three solutions contain a certain acid. The first contains 10% acid, the second 30%, and the third 50%. A chemist wishes to use all three solutions to obtain a 50-liter mixture containing 32% acid. If the chemist wants to use twice as much of the 50% solution as of the 30% solution, how many liters of each solution should be used?
Answers
Answer:
To prepare 50L of 32% solution you need: 11L of 30% solution, 22L of 50% solution and 17L of 10% solution.
Explanation:
A 32% solution of acid means 32L of acid per 100L of solution. As the chemist wants to make a solution using twice as much of the 50% solution as of the 30% solution it is possible to write:
2x*50% + x*30% + y*10% = 50L*32%
130x + 10y = 1600 (1)
-Where x are volume of 30% solution, 2x volume of 50% solution and y volume of 10% solution-
Also, it is possible to write a formula using the total volume (50L), thus:
2x + x +y = 50L
3x + y = 50L (2)
If you replace (2) in (1):
130x + 10(50-3x) = 1600
100x + 500 = 1600
100x = 1100
x = 11L -Volume of 30% solution-
2x = 22L -Volume of 50% solution-
50L - 22L - 11L = 17 L -Volume of 10% solution-
I hope it helps!
Final answer:
To create a 50-liter mixture with 32% acid concentration, the chemist should use 20 liters of the 10% acid solution, 10 liters of the 30% acid solution, and 20 liters of the 50% acid solution.
Explanation:
The question involves solving a system of linear equations to determine the volume of each acid solution required to create a 50-liter mixture with a 32% acid concentration. Let's denote the amount of the 10% solution as x liters, the 30% solution as y liters, and the 50% solution as 2y liters (since it's twice the amount of the 30% solution).
The total amount of acid should be 32% of the 50-liter mixture: 0.10x + 0.30y + 0.50(2y) = 0.32 × 50.
We've also established the relationship between the 30% and 50% solutions: 2y is twice the amount of y.
x + 3y = 50
0.10x + 0.30y + 1.00y = 16
Combining and rearranging gives us x = 20 liters, y = 10 liters, and thus 2y = 20 liters. Therefore, to obtain the desired mixture, the chemist should use 20 liters of the 10% solution, 10 liters of the 30% solution, and 20 liters of the 50% solution.
For this question please enter the number of sigma (\sigma σ ) and pi (\pi π ) bonds (e.g. 0,1,2,3,4, etc). How many sigma and pi bonds, respectively, are in this carboxylic acid? CH3CHCHCOOH. 11 \sigma σ bonds and 1 \pi π bond(s). How many sigma and pi bonds, respectively, are in this organic molecule (an amine)? HCCCH2CHCHNH2 \sigma σ bonds and \pi π bond(s). How many sigma and pi bonds, respectively, are in this organic molecule? H2CCHCH2CCH. \sigma σ bonds and \pi π bond(s).
Answers
Answer:
CH3CHCHCOOH : 11 sigma and 2 pi bonds
HCCCH2CHCHNH2: 12 sigma and 3 pi bonds
H2CCHCH2CCH: 10 sigma and 3 pi bonds
Explanation:
>In this carboxylic acid there are 11 sigma bonds between C-H, C-C ,C-O and O-H atoms,while there are two pi bonds ,one is made by carbonyl carbon and other from alkene C=C carbons
> In this amine there are 12 sigma bonds made by C-C,C-H and N-H atoms,while 2 pi bonds made by alkyne carbons C≡C carbons and 1 pi bond is made by alkene C=C carbon atoms
>In this organic molecule there are 10 sigma bonds made by C-C and C-H bonds ,while 2 pi bonds made by alkyne carbons C≡C carbons and 1 pi bond is made by alkene C=C carbon atoms.
Answer:
The number of sigma and pi bonds in (a) 11 sigma, 2 pi, (b) 12 sigma, 3 pi, (c) 10 sigma and 3 pi bonds.
Explanation:
Sigma bonds are formed by the linear interactions of the two atoms. The pi bonds are covalent bonds formed at top and bottom of the sigma bonds.
(a) [tex]\rm CH_3CHCHCOOH[/tex]
The compound is bonded with 7 Hydrogen with sigma bond. There is 3 sigma bond in between the 4 carbon atoms. One carbon atom is attached with OH group of carboxylic acid with 1 sigma bond. So, total, 11 Sigma bonds are present in the structure.
The pi bonds are 2 in number, 1 in between C-C in backbone. The pi bond is in carboxylic group in between C-O.
(b) [tex]\rm HCCCH_2CHCHNH_2[/tex]
There are 12 sigma bonds in the backbone of given amine. The number of pi bonds are 3. There are two pi bonds in between the alkyene carbons and one pi bond is present in the alkene carbon-carbon bond.
(c) [tex]\rm H_2CCHCH_2CCH[/tex]
The structure comprises of 10 sigma bonds in the backbone. There is presence of 3 pi bonds in the given compound structure.
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In the fermentation of glucose (wine making), 790 mL of CO2 gas was produced at 37 ∘C and 1.00 atm . You may want to reference (Pages 303 - 304) Section 8.5 while completing this problem. Part A What is the final volume, in liters, of the gas when measured at 18 ∘C and 695 mmHg , when n is constant?
Answers
Using the Combined Gas Law, the temperature is converted to Kelvin and pressure to atm, then the initial and final conditions are plugged into the formula to find that the final volume is approximately 0.808 liters.
The problem presented involves a gas law calculation where we need to find the final volume of the gas when measured at a different temperature and pressure while keeping the amount of gas constant. This is a straightforward application of the Combined Gas Law, which is stated as (P1 × V1) / T1 = (P2 × V2) / T2. Making sure to convert temperatures to Kelvin and pressures to atmospheres if they're not already can help in solving the problem accurately.
First, let's convert the temperatures from Celsius to Kelvin by adding 273.15. Therefore, 37 °C becomes 310.15 K and 18 °C becomes 291.15 K. Secondly, the pressure needs to be converted from mmHg to atm by dividing by 760. Thus, 695 mmHg becomes approximately 0.915 atm. Now we apply the Combined Gas Law:
P1 × (V1 / T1) = P2 × (V2 / T2)
(1.00 atm × 0.790 L) / 310.15 K = (0.915 atm × V2) / 291.15 K
After cross-multiplying and solving for V2, we find that the final volume of the gas when measured at 18°C and 695 mmHg, with a constant amount of gas, is approximately 0.808 liters.
4) Balance the following redox reaction in an acidic solution. What are the coefficients in front of H⁺ and Fe3+ in the balanced reaction? How many moles of electrons are transferred? Fe2+(aq) + MnO4⁻(aq) → Fe3+(aq) + Mn2+(aq)
Answers
Answer:
- The coefficients in front of H⁺ and Fe³⁺ are 8 and 5
- There are transferred 5 moles of e-
Explanation:
This is the reaction:
Fe²⁺(aq) + MnO₄⁻(aq) → Fe³⁺(aq) + Mn²⁺(aq)
Let's think the oxidation numbers:
Fe2+ changed to Fe3+
It has increased the oxidation number → OXIDATION
Mn in MnO₄⁻ acts with +7 and it changed to Mn²⁺
It has decreased the oxidation number → REDUCTION
Let's make the half reactions:
Fe²⁺ → Fe³⁺ + 1e⁻ (it has lost 1 mol of e⁻)
MnO₄⁻ + 5e⁻ → Mn²⁺ (it has gained 5 mol of e⁻)
Now we have to ballance the O. In acidic medium we complete with water as many oxygens we have, in the opposite side. We have 4 O in reactant side, so we fill up with 4 H2O in products side.
MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O
Now we have to ballance the H, so as we have 8 H in products side, we complete with 8H⁺ in reactants, this is the complete half reaction:
8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O
Notice that have 1e⁻ in oxidation and 5e⁻ in reduction. We must multiply (x5) the half reaction of oxidation, so the electrons can be cancelled.
(Fe²⁺ → Fe³⁺ + 1e⁻ ) .5
5Fe²⁺ → 5Fe³⁺ + 5e⁻
8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O
We sum both half reactions:
5Fe²⁺ + 8H⁺ + MnO₄⁻ + 5e⁻ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
The electrons are cancelled, so the ballanced reaction is this:
5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O
The given redox reaction balances to 5Fe2+(aq) + MnO4⁻(aq) + 8H⁺(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l), with coefficients of 8 for H⁺ and 5 for Fe3+ in the balanced reaction. A total of 5 moles of electrons are transferred.
Explanation:To balance the given redox reaction, you need to separate the two half-reactions, one for oxidation and one for reduction. In the case of Fe2+(aq) + MnO4⁻(aq) → Fe3+(aq) + Mn2+(aq), the oxidation half-reaction is Fe2+(aq) → Fe3+(aq) + e⁻, and the reduction half-reaction is MnO4⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn2+(aq) + 4H2O(l).
After balancing the half-reactions, you can add them together to get 5Fe2+(aq) + MnO4⁻(aq) + 8H⁺(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l). This reaction shows that the coefficients in front of H⁺ and Fe3+ in the balanced reaction are 8 and 5, respectively. Also, the stoichiometric coefficients in front of the electrons in the two half-reactions indicate that a total of 5 moles of electrons are transferred during the reaction.
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Draw the Lewis structure for each of the following and then determine if the molecule is polar or nonpolar. Do not draw double bonds to oxygen atoms unless they are needed for the central atom to obey the octet rule. Do not include formal charges in your drawing. NO2Cl.
Answers
NO2Cl is a polar molecule because the electronegativity difference between atoms of the molecule, being the chloride tho most electronegative. This fact and the asymmetry of the molecule makes a dipole moment.
In order to don’t include formal charges in the drawing you can use the image number 1. Nevertheless is more usual to use a lewis structure like image number 2.
Given thatH2(g) + F2(g) -> 2HF(g) => ∆H = -546.6 kJ . mol-12H2(g) + O2(g) -> 2H20(l) => ∆H = -571.6 kJ. mol-1
Calculate the value of ∆H for:2F2(g) + 2H20(l) -> 4HF(g) + O2(g)
Answers
Answer: The [tex]\Delta H^o_{rxn}[/tex] for the reaction is -521.6 kJ.
Explanation:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical equation for the reaction of fluorine and water follows:
[tex]2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)[/tex] [tex]\Delta H^o_{rxn}=?[/tex]
The intermediate balanced chemical reaction are:
(1) [tex]H_2(g)+F_2(g)\rightarrow 2HF(g)[/tex] [tex]\Delta H_1=-546.6kJ[/tex] ( × 2)
(2) [tex]H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex] [tex]\Delta H_2=-571.6kJ[/tex]
The expression for enthalpy of reaction follows:
[tex]\Delta H^o_{rxn}=[2\times \Delta H_1]+[1\times (-\Delta H_2)][/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(2\times (-546.6))+(1\times (571.6))]=-521.6kJ[/tex]
Hence, the [tex]\Delta H^o_{rxn}[/tex] for the reaction is -521.6 kJ.
Given the equation C3H8(g) + O2(g) = CO2(g) + H2O(g) and that the enthalpies of formation for H2O(g) = -241.8 kJ/mol, CO2(g) = -393.5kJ/mol, and the enthalpy of combustion for the reaction is -2220.1kJ/mol, what is the heat of formation of propane?
Answers
Answer: 72.4 kJ/mol
Explanation:
The balanced chemical reaction is,
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)[/tex] [tex]\Delta H=-2220.1kJ/mol[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})][/tex]
where,
n = number of moles
[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
[tex]-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})][/tex]
[tex]\Delta H_{C_3H_8}=72.4kJ/mol[/tex]
Therefore, the heat of formation of propane is 72.4 kJ/mol
The heat of formation of propane can be found using Hess's Law and the given enthalpy formations of CO2 and H2O, along with the enthalpy combustion of propane. Calculations reveal a heat of formation for propane of -103.85 kJ/mol.
Explanation:The heat of formation of propane can be calculated using Hess's Law, which states that the enthalpy change in a chemical reaction is independent of the pathway between the initial and final states. In this case, we know the enthalpy of combustion for propane (C3H8) is -2220.1kJ/mol, and that the standard enthalpy of formation (ΔHf°) for CO2 and H2O are -393.5kJ/mol and -241.8kJ/mol, respectively.
Therefore, by rearranging the equation ΔH(combustion) = Σ(ΔHf° products) - Σ(ΔHf° reactants), we can solve for the ΔHf° of propane.
Inserting the given values into the equation, -2220.1 kJ/mol = [ 3(-393.5 kJ/mol) + 4(-241.8 kJ/mol)] - [x + 5(0)], where x represents the ΔHf° of propane, and '5(0)' is the contribution from the oxygen, which is zero because the change in enthalpy formation for an element in its stable form is zero. Thus, solving for x, you get a heat of formation for propane of -103.85 kJ/mol.
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