Problem: Use data from Appendix IIB in the textbook to calculate the equilibrium constants at 25 ˚C for each of the following reactions.2 CO(g) + O2(g) ⇌ 2 CO2(g)

FREE Expert Solution

Recall that ΔG˚rxn and K are related to each other:


ΔG°rxn=-RTlnK


We can use the following equation to solve for ΔG˚rxn:


ΔG°rxn=ΔH°rxn-TΔS°rxn



Step 1: We can use the following equation to solve for ΔH˚rxn:


ΔH°rxn=ΔH°f, prod-ΔH°f, react


Note that we need to multiply each ΔH˚f by the stoichiometric coefficient since ΔH˚f is in kJ/mol. 

Also, note that ΔH˚f for elements in their standard state is 0.


We also need to convert ΔH˚rxn from kJ/mol to J/ mol so our units remain consistent.


ΔH°rxn=(2 mol CO2)-393.5 kJ1 mol CO2           -(2 mol CO)-99 kJ1 mol CO+0

ΔH°rxn=-589 kJmol×103 J1 kJ

ΔH˚rxn = –5.89 × 105 J/mol



Step 2: We can use the following equation to solve for ΔS˚rxn:


ΔS°rxn=S°f, prod-S°f, react


Note that we need to multiply each S˚ by the stoichiometric coefficient since S˚ is in J/mol • K.


ΔS°rxn=2 mol CO251.07 Jmol·K           -2 mol CO197.66 Jmol·K+1 mol O2205mol·K

ΔS˚rxn = –498.18 J/mol • K



Step 3: Now that we have ΔH˚rxn and ΔS˚rxn, we can now solve for ΔG˚rxn

The given temperature is 298 K.


ΔG°rxn=ΔH°rxn-TΔS°rxn

ΔG°rxn=-5.89×105 Jmol-(298 K)-498.18Jmol·K

ΔG˚rxn = –440542.36 J/mol



Step 4: Solving for K:


ΔG°=-RTlnK

-440542.36 Jmol=-8.314Jmol·K(298 K)(ln K)

lnK=-440542.36 Jmol-2477.57 Jmol=177.81

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Problem Details

Use data from Appendix IIB in the textbook to calculate the equilibrium constants at 25 ˚C for each of the following reactions.

CO(g) + O2(g) ⇌ 2 CO2(g)

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