Question

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).

Answer 1

Answer 2

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