Answer:
I can provide the solutions to the given problems:
1. ΔG = ΔH - TΔS
ΔG = (-176.0 kJ/mol) - (298 K)(-0.2848 kJ/K mol)
ΔG = -176.0 kJ/mol + 85.04 kJ/mol
ΔG = -90.96 kJ/mol
The reaction will occur spontaneously as ΔG is negative.
2. ΔG = ΔH - TΔS
ΔG = (-2 mol)(0 kJ/mol) - [(1 mol)(191.5 kJ/mol) + (3 mol)(0 kJ/mol) - (2 mol)(130.6 kJ/mol)]
ΔG = -191.5 kJ/mol + 261.2 kJ/mol
ΔG = 69.7 kJ/mol
The reaction is non-spontaneous as ΔG is positive.
3. ΔH = ΣΔH(products) - ΣΔH(reactants)
ΔH = (2 mol)(-296.8 kJ/mol) + (-395.5 kJ/mol)
ΔH = -989.1 kJ/mol
ΔS = ΣS(products) - ΣS(reactants)
ΔS = (2 mol)(205.2 J/K mol) + (1 mol)(205.0 J/K mol) - (2 mol)(248.0 J/K mol) - (1 mol)(298.0 J/K mol)
ΔS = 164 J/K mol
ΔG = ΔH - TΔS
ΔG = (-989.1 kJ/mol) - (523 K)(-0.164 kJ/K mol)
ΔG = -989.1 kJ/mol + 85.372 kJ/mol
ΔG = -903.7 kJ/mol
4. ΔG°f for 2CO2(g) = (-394 kJ/mol)(2) = -788 kJ/mol
ΔG°f for 4H2O(g) = (-237 kJ/mol)(4) = -948 kJ/mol
ΔG°f for 2CH3OH(g) = (-163 kJ/mol)(2) = -326 kJ/mol
ΔG°f for 3O2(g) = (-229 kJ/mol)(3) = -687 kJ/mol
ΔG = ΣΔG°f(products) - ΣΔG°f(reactants)
ΔG = (-788 kJ/mol - 948 kJ/mol) - [(-326 kJ/mol) + (-687 kJ/mol)]
ΔG = -1733 kJ/mol + 1013 kJ/mol
ΔG = -720 kJ/mol
5. ΔG° = ΣΔG°f(products) - ΣΔG°f(reactants)
ΔG° = [2(-270 kJ/mol) + 1/2(0 kJ/mol)] - (1(-502 kJ/mol))
ΔG° = -40 kJ/mol
The reaction will occur spontaneously as ΔG° is negative.
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Verified answer
Answer:
I can provide the solutions to the given problems:
1. ΔG = ΔH - TΔS
ΔG = (-176.0 kJ/mol) - (298 K)(-0.2848 kJ/K mol)
ΔG = -176.0 kJ/mol + 85.04 kJ/mol
ΔG = -90.96 kJ/mol
The reaction will occur spontaneously as ΔG is negative.
2. ΔG = ΔH - TΔS
ΔG = (-2 mol)(0 kJ/mol) - [(1 mol)(191.5 kJ/mol) + (3 mol)(0 kJ/mol) - (2 mol)(130.6 kJ/mol)]
ΔG = -191.5 kJ/mol + 261.2 kJ/mol
ΔG = 69.7 kJ/mol
The reaction is non-spontaneous as ΔG is positive.
3. ΔH = ΣΔH(products) - ΣΔH(reactants)
ΔH = (2 mol)(-296.8 kJ/mol) + (-395.5 kJ/mol)
ΔH = -989.1 kJ/mol
ΔS = ΣS(products) - ΣS(reactants)
ΔS = (2 mol)(205.2 J/K mol) + (1 mol)(205.0 J/K mol) - (2 mol)(248.0 J/K mol) - (1 mol)(298.0 J/K mol)
ΔS = 164 J/K mol
ΔG = ΔH - TΔS
ΔG = (-989.1 kJ/mol) - (523 K)(-0.164 kJ/K mol)
ΔG = -989.1 kJ/mol + 85.372 kJ/mol
ΔG = -903.7 kJ/mol
4. ΔG°f for 2CO2(g) = (-394 kJ/mol)(2) = -788 kJ/mol
ΔG°f for 4H2O(g) = (-237 kJ/mol)(4) = -948 kJ/mol
ΔG°f for 2CH3OH(g) = (-163 kJ/mol)(2) = -326 kJ/mol
ΔG°f for 3O2(g) = (-229 kJ/mol)(3) = -687 kJ/mol
ΔG = ΣΔG°f(products) - ΣΔG°f(reactants)
ΔG = (-788 kJ/mol - 948 kJ/mol) - [(-326 kJ/mol) + (-687 kJ/mol)]
ΔG = -1733 kJ/mol + 1013 kJ/mol
ΔG = -720 kJ/mol
5. ΔG° = ΣΔG°f(products) - ΣΔG°f(reactants)
ΔG° = [2(-270 kJ/mol) + 1/2(0 kJ/mol)] - (1(-502 kJ/mol))
ΔG° = -40 kJ/mol
The reaction will occur spontaneously as ΔG° is negative.