The conversion of n-butane to 2-methylpropane is an equilibrium process with ΔH° = −2.05 kcal/mol and ΔG° = −0.89 kcal/mol at 298 K. What is the equilibrium constant for this reaction?
Added by Maddi H.
Step 1
Given ΔH° = -2.05 kcal/mol, ΔG° = -0.89 kcal/mol, and T = 298 K. ΔS = (-0.89 kcal/mol - (-2.05 kcal/mol)) / 298 K ΔS = 1.16 kcal/mol / 298 K ΔS = 0.0039 kcal/(mol*K) Show more…
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