Consider the chemical reaction and thermochemical information
and initial partial pressures of the reaction components given
below:
CH3OH(g) + HBr(g) ⇌ CH3Br(g) +
H2O(g)
ΔHf° (kJ/mol) | S° (J mol-1 K-1) | P (atm) | |||
CH3Br | -35.50 | 246.40 | 0.462 | ||
H2O | -241.83 | 188.84 | 0.984 | ||
CH3OH | -201.50 | 239.80 | 5.82 | ||
HBr | -36.29 | 198.70 | 5.22 |
Determine ΔG (in kJ) for this reaction at 867.29 K. Assume
ΔH°f and S° do not vary as a function of temperature.
Report your answer to two decimal places in standard notation (i.e.
123.45 kJ).
To determine ΔG (Gibbs free energy) for the given reaction at 867.29 K, we can use the equation:
ΔG = ΔH - TΔS
where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.
Given: ΔHf° (kJ/mol) and S° (J mol-1 K-1) for each component: CH3Br: ΔHf° = -35.50 kJ/mol, S° = 246.40 J mol-1 K-1 H2O: ΔHf° = -241.83 kJ/mol, S° = 188.84 J mol-1 K-1 CH3OH: ΔHf° = -201.50 kJ/mol, S° = 239.80 J mol-1 K-1 HBr: ΔHf° = -36.29 kJ/mol, S° = 198.70 J mol-1 K-1
The reaction can be written as: CH3OH(g) + HBr(g) ⇌ CH3Br(g) + H2O(g)
Now, let's calculate ΔG for the reaction using the given information:
ΔG = ΣΔGf°(products) - ΣΔGf°(reactants)
First, let's calculate the ΔGf° (standard Gibbs free energy change of formation) for each component:
ΔGf° (CH3Br) = ΔHf° (CH3Br) - TΔS° (CH3Br) = (-35.50 kJ/mol) - (867.29 K * (246.40 J/mol K / 1000)) = -35.50 kJ/mol - 214.19 kJ/mol = -249.69 kJ/mol
ΔGf° (H2O) = ΔHf° (H2O) - TΔS° (H2O) = (-241.83 kJ/mol) - (867.29 K * (188.84 J/mol K / 1000)) = -241.83 kJ/mol - 163.67 kJ/mol = -405.50 kJ/mol
ΔGf° (CH3OH) = ΔHf° (CH3OH) - TΔS° (CH3OH) = (-201.50 kJ/mol) - (867.29 K * (239.80 J/mol K / 1000)) = -201.50 kJ/mol - 207.06 kJ/mol = -408.56 kJ/mol
ΔGf° (HBr) = ΔHf° (HBr) - TΔS° (HBr) = (-36.29 kJ/mol) - (867.29 K * (198.70 J/mol K / 1000)) = -36.29 kJ/mol - 172.18 kJ/mol = -208.47 kJ/mol
Now, let's substitute these values into the equation to calculate ΔG:
ΔG = ΣΔGf°(products) - ΣΔGf°(reactants) = [ΔGf° (CH3Br) + ΔGf° (H2O)] - [ΔGf° (CH3OH) + ΔGf° (HBr)] = [-249.69 kJ/mol + (-405.50 kJ/mol)] - [-408.56 kJ/mol + (-208.47 kJ/mol)] = -655.19 kJ/mol + 616.03 kJ/mol = -39.16 kJ/mol
Consider the chemical reaction and thermochemical information and initial partial pressures of the reaction components given...
For the reaction below, determine the equilibrium partial pressure (in atm) of H2 at 714.10 K if the initial pressure of HBr is 1.24 atm. Assume that ΔH and ΔS do not vary with temperature. Report your answer to three significant figures. LiH (s) + HBr (g) ⇌ LiBr (s) + H2 (g) ΔHf° (kJ/mol) S° (J mol-1 K-1) LiBr -351.20 74.30 H2 0.00 130.68 LiH -90.50 20.00 HBr -36.29 198.70
Use the provided thermodynamic data at 298 K and partial pressures given to decide in which direction (forward or backward) the reaction will proceed under these conditions: Compound ΔG (kJ/mol) Partial Pressure (atm) SO2 (g) -300.2 1.0 x 10-4 O2 (g) 0.0 0.20 SO3 (g) -371.1 0.10
Calculate ΔH for the following reaction, CaO(s) + CO2(g) → CaCO3(s) given the thermochemical equations below. 2 Ca(s) + O2(g) → 2 CaO(s) ΔH = -1270.2 kJ C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ 2 Ca(s) + 2 C(s) + 3 O2(g) → 2 CaCO3(s) ΔH = -2413.8 kJ A compound contains C, H and O as the elements. A 20.0 g-sample is comprised of 1.34 g H and also 8.00 g of C. What...
Use the given data at 700 K to calculate ΔG°for the reaction H2(g) + CO2(g) → H2O(g) + CO(g) Substance H2(g) CO2(g) H2O(g) CO(g) ΔH°f(kJ/mol) 0 -393 -242 -111 S°(J/K·mol) 131 214 189 198 -472 kJ 10.6 kJ -775 kJ -2.94 x 104 kJ
Use the given data at 500 K to calculate ΔG°for the reaction 2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) Substance H2S(g) O2(g) H2O(g) SO2(g) ΔH°f(kJ/mol) -21 0 -242 -296.8 S°(J/K·mol) 206 205 189 248
7 Consider the following reaction at 298 K: NH3(g) +HCI(g)NH4CI(s) Using the thermochemical data at 298 K given below (a) Determine the AH for the reaction. (2 marks) (b) Determine the AG for the reaction and comment on its value. (3 marks) (c) Determine the maximum work done available in the reaction. (3 marks) An equilibrium mixture, consisting of 0.5 atm of NHy(), 0.5 atm of HCKg) and excessive of NH Cl(s), is enclosed in a container at 298 K...
I wasn't given the partial pressures so why do you need it? can the problem really not be solved without it? 1 bar for each 4. (11.11) The combustion of hydrogen is a reaction that is known to "go to completion." a. Use data in Appendix H to evaluate the thermodynamic equilibrium constant at 298.15 K for the reaction H2(g) + {O2(g) → H2O(1) b. Assume that the reaction is at equilibrium at 298.15 K in a system in which...
1. The initial concentrations of reactants and products for this reaction are given below. N2(g) + O2(g) ⇄ 2NO(g) Calculate Q for this reaction. Answer this to one decimal place (e.g. 10.2) The initial concentration of N2 is 1.0 M The initial concentration of O2 is 1.0 M The initial concentration of NO is 2.5 M 2. The reaction below is not at equilibria and Q = 1.7. The equilibrium constant is K = 0.230. 2SO3(g) ⇌ 2SO2(g) + O2(g)...