Problem: Calculate K at 298 K for the following reaction: NO(g) + 1/2 O2 (g) ⇌ NO2(g) ΔG (kJ/mol) for the following: NO(g) 86.60, NO2(g) 51, O2(g) 0 (answer in scientific notation)

FREE Expert Solution

NO(g) + 1/2 O2 (g) ⇌ NO2(g) 


ΔG°rxn=-RTlnK



Step 1: Calculate for ΔG˚rxn:


ΔG°rxn=ΔG°f, prod-ΔG°f, react


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


ΔG°rxn=(1 mol NO2)51 kJ1 mol NO2           -(1 mol NO)86.60 kJ1 mol NO+0

ΔG°rxn=-35.6 kJmol×103 J1 kJ

ΔG˚rxn = –35600 J/mol



Step 2: Calculate K:

R = 8.314 J/mol • K

T = 298 K


Solving for K:

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Problem Details

Calculate K at 298 K for the following reaction: NO(g) + 1/2 O2 (g) ⇌ NO2(g) 

ΔG (kJ/mol) for the following: 

NO(g) 86.60, NO2(g) 51, O2(g) 0 (answer in scientific notation)