Problem: 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) = 49ΔGf° (H2O(l)) = -241ΔGf° (HNO3) = -84ΔGf° (NO) = 83

FREE Expert Solution

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:


ΔG°rxn=ΔG°f, prod-ΔG°f, react


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Problem Details

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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