We’re being asked to determine the ΔG° in kJ for the reaction at 298 K
3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)
We can use the following equation to solve for ΔG˚rxn:
Given the values of ΔGf° given below in kJ/mol, calculate the value of ΔG° in kJ for the reaction at 298 K.:
3 NO2(g) + H2O(l) => 2 HNO3(l) + NO(g)
ΔGf° (NO2) = 49
ΔGf° (H2O(l)) = -241
ΔGf° (HNO3) = -84
ΔGf° (NO) = 83
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