Answer:
The difference in energy between the [tex]S_1[/tex] and [tex]T_1[/tex] states is 35.7 kJ/mol.
Explanation:
The Planck's equation :
[tex]E=\frac{hc}{\lambda} [/tex]
Where:
E = Energy of the electromagnetic radiations.
h = Planck's constant = [tex]6.626\times 10^{-34} Js[/tex]
c = speed of light = [tex]3\times 10^8 m/s[/tex]
[tex]\lambda [/tex] = Wavelength of the electromagnetic radiations.
Energy associated with [tex]T_1[/tex] transition : [tex]E_1[/tex]
Wavelength associated with [tex]T_1[/tex] transition :
[tex]\lambda _1=397 nm=397\times 10^{-9} m[/tex]
Energy associated with [tex]S_1[/tex] transition : [tex]E_2[/tex]
Wavelength associated with [tex]S_1[/tex] transition : [tex]\lambda _2=355 nm=355\times 10^{-9} m[/tex]
The difference in energy between the [tex]S_1[/tex] and [tex]T_1[/tex] states:
[tex]E=E_1-E_2=\frac{hc}{\lambda _2}-\frac{hc}{\lambda _1}[/tex]
[tex]E=hc\times (\frac{1}{\lambda _2}-\frac{1}{\lambda _1})[/tex]
[tex]E=6.626\times 10^{-34} Js \times 3\times 10^8 m/s(\frac{1}{355\times 10^{-9}m}-\frac{1}{397\times 10^{-9} m})[/tex]
[tex]E=5.92\times 10^{-20} J[/tex]
1 J = 0.001 kJ
[tex]E=5.92\times 10^{-20}\times 10^{-3} kJ=5.92\times 10^{-23} kJ[/tex]
1 mole = [tex]6.022\times 10^{-23}[/tex]
The difference in energy(kJ/mol) between the [tex]S_1[/tex] and [tex]T_1[/tex] states:
[tex]=5.92\times 10^{-23} kJ\times 6.022\times 10^{23} mol^{-1}=35.7 kJ/mol[/tex]
Final answer:
The energy difference between the S1 and T1 states in formaldehyde can be calculated using the wavelengths given and Planck's equation, with the resulting value representing the difference in energy due to electronic transitions between these two states.
Explanation:
The student is asking about the difference in energy between the singlet state (S1) and triplet state (T1) of formaldehyde, which corresponds to n → π* electronic transitions occurring at different wavelengths. The wavelengths given are 397 nm (T1) and 355 nm (S1). We can calculate the energy difference using the equation E = hv (where h is Planck constant and v is frequency), and converting wavelengths to frequencies using c = λv (where c is the speed of light and λ is the wavelength). To find the difference in energy (ΔE) in kJ/mol, we first convert the energy of each transition from Joules to kJ/mol and then subtract the energy of T1 from S1.
To calculate the energy for each wavelength:
E(397nm) = (6.626 x 10^-34 J·s) (3 x 10^8 m/s) / (397nm x 10^-9 m/nm) x (1/1000) kJ/J x (6.022 x 10^23 mol^-1)
E(355nm) = (6.626 x 10^-34 J·s) (3 x 10^8 m/s) / (355nm x 10^-9 m/nm) x (1/1000) kJ/J x (6.022 x 10^23 mol^-1)
ΔE = E(355nm) - E(397nm)
The difference in energy between the S_1 and T_1 states:
E=E_1-E_2=(hc)/(\lambda _2)-(hc)/(\lambda _1)
E=hc* ((1)/(\lambda _2)-(1)/(\lambda _1))
E=6.626* 10^(-34) Js * 3* 10^8 m/s((1)/(355* 10^(-9)m)-(1)/(397* 10^(-9) m))
E=5.92* 10^(-20) J
1 J = 0.001 kJ
E=5.92* 10^(-20)* 10^(-3) kJ=5.92* 10^(-23) kJ
1 mole = 6.022* 10^(-23)
The difference in energy(kJ/mol) between the S_1 and T_1 states:
=5.92* 10^(-23) kJ* 6.022* 10^(23) mol^(-1)=35.7 kJ/mol
The balanced combustion reaction for C 6 H 6 C6H6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 7.300 g C 6 H 6 7.300 g C6H6 is burned and the heat produced from the burning is added to 5691 g 5691 g of water at 21 ∘ 21 ∘ C, what is the final temperature of the water?
Answer: The final temperature of the water is [tex]33.85^{o}C[/tex].
Explanation:
We know that molar mass of [tex]C_{6}H_{6}[/tex] is 78 g/mol. And, the amount of heat produced when 2 mol of [tex]C_{2}H_{6}[/tex] burns is 6542 KJ.
This means that,
[tex]78 \times 2[/tex] = 156 g of [tex]C_{2}H_{6}[/tex] burns, heat produced is 6542 kJ.
Therefore, heat produced (Q) by burning 7.3 g of [tex]C_{6}H_{6}[/tex] is as follows.
[tex]\frac{6542 \times 7.3 g}{156 g}[/tex]
= 306.13 kJ
or, = 306130 J (as 1 KJ = 1000 J)
For water, mass is given as 5691 g and specific heat capacity of water is 4.186 [tex]J/g^{o}C[/tex].
So, we will calculate the value of final temperature as follows.
Q = [tex]m \times C \times (T_{f} - T_{i})[/tex]
306130 J = [tex]5691 g \times 4.186 J/g^{o}C \times (T_{f} - 21)^{o}C[/tex]
[tex](T_{f} - 21)^{o}C = \frac{306130 J}{23822.53 J/^{o}C}[/tex]
[tex]T_{f}[/tex] = 12.85 + 21
= [tex]33.85^{o}C[/tex]
Thus, we can conclude that the final temperature of the water is [tex]33.85^{o}C[/tex].
Final answer:
To determine the final temperature of water when 7.300 g of benzene is burned, use the enthalpy of combustion for benzene and apply stoichiometry to find the heat transferred to the water, then use the specific heat capacity formula to find the change in water's temperature.
Explanation:
The question asks about finding the final temperature of water after burning 7.300 g of benzene (C6H6) and adding the heat produced to the water at an initial temperature of 21 °C. To answer this, we need to use thermochemical principles and calculate the energy involved in the combustion of benzene which will then be absorbed by the water, causing a change in its temperature.
The heat absorbed by the water (q_water) can be calculated using the formula q = m * c *
igtriangleup T, where m is the mass of the water, c is the specific heat capacity of water (4.184 J/g°C), and
igtriangleup T is the change in temperature. Since we're given the enthalpy of combustion of benzene (6,542 kJ per 2 moles of C6H6), we can calculate the heat produced by burning 7.300 g of benzene using stoichiometry and the molar mass of C6H6 (78.11 g/mol).
Using this information allows us to solve for the final temperature of the water. Remember to convert the energy units appropriately when performing these calculations.
Calculate the standard free energy change for the combustion of one mole of methane using the values for standard free energies of formation of the products and reactants. The sign of the standard free energy change allows chemists to predict if the reaction is spontaneous or not under standard conditions and whether it is product-favored or reactant-favored at equilibrium.
Answer:
The standard free energy of combustion of 1 mole of methane = -801.11 kJ
The negative sign shows that this reaction is spontaneous under standard conditions.
The negative sign on the standard free energy also means this combustion reaction is product-favoured at equilibrium.
Explanation:
The chemical reaction for the combustion of methane is given by
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
The standard free energy of formation for the reactants and products as obtained from literature include
For CH₄, ΔG⁰ = -50.50 kJ/mol
For O₂, ΔG⁰ = 0 kJ/mol
For CO₂, ΔG⁰ = -394.39 kJ/mol
For H₂O(g), ΔG⁰ = -228.61 kJ/mol
ΔG(combustion) = ΔG(products) - ΔG(reactants)
ΔG(products) = (1×-394.39) + (2×-228.61) = -851.61 kJ/mol
ΔG(reactants) = (1×-50.50) + (2×0) = -50.50 kJ/mol
ΔG(combustion) = ΔG(products) - ΔG(reactants)
ΔG(combustion) = -851.61 - (-50.5) = -801.11 kJ/mol
Since we're calculating for 1 mole of methane, ΔG(combustion) = -801.11 kJ
- A negative sign on the standard free energy means that the reaction is spontaneous under standard conditions.
- A positive sign indicates a non-spontaneous reaction.
- A Gibb's free energy of 0 indicates that the reaction is at equilibrium.
- A negative sign on the standard free energy also means that if the reaction reaches equilibrium, it will be product favoured.
- A positive sign on the standard free energy means that the reaction is reactant-favoured at equilibrium.
Hope this Helps!!!
Methanol and oxygen react to form carbon dioxide and water, like this: 2CH3OH (l) + 3O2 (g) → 2CO2 (g) + 4H2O (g) At a certain temperature, a chemist finds that a 9.3L reaction vessel containing a mixture of methanol, oxygen, carbon dioxide, and water at equilibrium has the following composition:compound CH3OH O2 CO2 H2Oamount 1.47g 1.56g 2.28g 3.33gCalculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.
Answer:
The value of the equilibrium constant Kc for this reaction is 3.34*10^-5
Explanation:
Step 1: Data given
Volume of the reaction vessel = 9.3 L
Mass of CH3OH at equilibrium = 1.47 grams
Mass of O2 at equilibrium = 1.56 grams
Mass of CO2 at equilibrium = 2.28 grams
Mass of H2O at equilibrium = 3.33 grams
Molar mass CH3OH = 32.04 g/mol
Molar mass of O2 = 32.0 g/mol
Molar mass of CO2 = 44.01 g/mol
Molar mass of H2O = 18.02 g/mol
Step 2: The balanced equation
2CH3OH (l) + 3O2 (g) → 2CO2 (g) + 4H2O (g)
Step 3: Calulate moles
Moles = mass / molar mass
Moles CH3OH = 1.47 grams / 32.04 g/mol
Moles CH3OH = 0.0459 moles
Moles O2 = 1.56 grams / 32.0 g/mol
Moles O2 = 0.0489 moles
Moles CO2 = 2.28 grams / 44.01 g/mol
Moles CO2 = 0.0518 moles
Moles H2O = 3.33 grams / 18.02 g/mol
Moles H2O = 0.185 moles
Step 4: Calculate molarity
Molarity = moles / volume
[CH3OH] = 0.0459 moles / 9.3 L
[CH3OH] = 0.00494 M
[O2] = 0.0489 moles / 9.3 L
[O2] = 0.00526 M
[CO2] = 0.0518 moles / 9.3 L
[CO2] = 0.00557 M
[H2O] = 0.185 moles / 9.3 L
[H2O] = 0.0199 M
Step 5: Calculate Kc
Since liquids and solids do not affect the equilibrium, CH3OH will not take part
Kc = [CO2]²[H2O]^4 / [O2]³
Kc = (0.00557² * 0.0199^4) / 0.00526³
Kc = 3.34*10^-5
The value of the equilibrium constant Kc for this reaction is 3.34*10^-5
The value of the equilibrium constant [tex]\( K_c \)[/tex] for this reaction is approximately 1.4.
To calculate the equilibrium constant [tex]\( K_c \)[/tex], we need to determine the molar concentrations of each species involved in the reaction at equilibrium. The balanced chemical equation is:
[tex]\[ 2CH_3OH (l) + 3O_2 (g) \rightleftharpoons 2CO_2 (g) + 4H_2O (g) \][/tex]
Given the masses of the compounds and the volume of the reaction vessel, we can calculate the molar concentrations as follows:
1. Calculate the moles of each compound:
[tex]- For \( CH_3OH \): \( n_{CH_3OH} = \frac{1.47 \text{ g}}{32.04 \text{ g/mol}} \)\\ - For \( O_2 \): \( n_{O_2} = \frac{1.56 \text{ g}}{32.00 \text{ g/mol}} \)\\ - For \( CO_2 \): \( n_{CO_2} = \frac{2.28 \text{ g}}{44.01 \text{ g/mol}} \)\\ - For \( H_2O \): \( n_{H_2O} = \frac{3.33 \text{ g}}{18.015 \text{ g/mol}} \)[/tex]
2. Calculate the molar concentrations (using the volume of the reaction vessel, 9.3 L):
[tex]- For \( CH_3OH \): \( [CH_3OH] = \frac{n_{CH_3OH}}{9.3 \text{ L}} \)\\ - For \( O_2 \): \( [O_2] = \frac{n_{O_2}}{9.3 \text{ L}} \)\\ - For \( CO_2 \): \( [CO_2] = \frac{n_{CO_2}}{9.3 \text{ L}} \)\\ - For \( H_2O \): \( [H_2O] = \frac{n_{H_2O}}{9.3 \text{ L}} \)[/tex]
3. Write the expression for [tex]\( K_c \)[/tex] and substitute the concentrations:
[tex]- \( K_c = \frac{[CO_2]^2 [H_2O]^4}{[CH_3OH]^2 [O_2]^3} \)[/tex]
4. Substitute the calculated molar concentrations into the expression for [tex]\( K_c \)[/tex] and solve.
5. Round the result to two significant digits.
After performing these calculations, we find that [tex]\( K_c \)[/tex] is approximately 1.4 when rounded to two significant digits. This value indicates the position of equilibrium in this reaction mixture, with higher values of [tex]\( K_c \)[/tex]corresponding to a greater extent of the reaction proceeding to the right (towards the products).
What to fill in the blanks with?
choices are: a shorter carbon chain, water, a longer carbon chain, other alcohol molecules, CH3 groups, ionic bonds, covalent bonds, a stronger interaction between CH3 groups, and hydrogen bonds.
1. 1-propanol can form __________ with _______, but butane cannot.
2. 1-propanol is more soluble than ethyl methyl ether because it can form more _______ with _______.
3. Ethanol is more soluble in water than 1-hexanol because it has ______.
Answer: 1. 1-propanol can form hydrogen bonds with water, but butane cannot.
2. 1-propanol is more soluble than ethyl methyl ether because it can form more hydrogen bonds with water.
3. Ethanol is more soluble in water than 1-hexanol because it has a shorter carbon chain.
Explanation:
It is know that like dissolves like. That is polar substance will be soluble in polar solvent and non-polar substance will be soluble in non-polar solvent.
As 1-propanol is polar in nature due to the alcoholic group present in it so, it is able to form hydrogen bonds with water. Hence, it readily dissolves in water. Whereas butane is non-polar in nature and it will not dissolve in water (polar solvent).
Also, smaller is the number of carbon chain present in a compound smaller will be its surface area. Hence, more readily it will dissolve in water.
For example, ethanol has less number of carbon atoms than 1-hexanol. So, ethanol will be more soluble in water than 1-hexanol.
Thus, we can conclude that given blanks are filled as follows.
1. 1-propanol can form hydrogen bonds with water, but butane cannot.
2. 1-propanol is more soluble than ethyl methyl ether because it can form more hydrogen bonds with water.
3. Ethanol is more soluble in water than 1-hexanol because it has a shorter carbon chain.
Manganese metal can be prepared by the thermite process: 4Al(s) . 3MnO2(s). -->. 3Mn(l). 2Al2O3(s) molar masses, in g/mol: 26.98. 86.94 54.94 101.96 If 203 g of Al and 472 g of MnO2 are mixed, which is the limiting reactant?
Answer:
MnO2 is the limiting reactant
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
4Al(s) + 3MnO2(s) —> 3Mn(l) + 2Al2O3(s)
Step 2:
Determination of the masses of Al and MnO2 that reacted from the balanced equation. This is illustrated below:
Molar Mass of Al = 26.98g/mol
Mass of Al from the balanced equation = 4 x 26.98 = 107.92g
Molar Mass of MnO2 = 86.94g/mol
Mass of MnO2 from the balanced equation = 3 x 86.94 = 260.82g
From the balanced equation above,
107.92g of Al reacted with 260.82g of MnO2 reacted.
Step 3:
Determination of the limiting reactant. This is illustrated below:
Let us consider using all the 203g of Al given to verify if there will be any leftover for MnO2. This is illustrated below:
From the balanced equation above,
107.92g of Al reacted with 260.82g of MnO2,
Therefore, 203g of Al will react with = (203 x 260.82)/107.92 = 490.61g of MnO2.
We can see that the mass of MnO2 that will react with 203g of Al is 490.61g which far greater than the 472g that was given. Therefore, it is not acceptable.
Now let us consider using all 472g of MnO2 given to verify if there will be any leftover for Al. This is shown below:
From the balanced equation above,
107.92g of Al reacted with 260.82g of MnO2,
Therefore, Xg of Al will react with 472g of MnO2 i.e
Xg of Al = (107.92 x 472)/260.82
Xg of Al = 195.30g
From the calculations above, we can see clearly that there are leftover for Al as 195.30g out of 203g reacted.
Therefore, MnO2 is the limiting reactant