Q.When 1-pentyne (A) is treated with 4N alcoholic KOH at $175^\circ$C, it is converted slowly into an equilibrium mixture of 1.3% 1-pentyne(A), 95.2% 2-pentyne(B) and 3.5% of 1,2 pentadiene (C). The equilibrium was maintained at $175^\circ$C, calculate $\Delta G^0$ for the following equilibria. $B \rightleftharpoons A$, $\Delta G_1^0 = ?$; $B \rightleftharpoons C$, $\Delta G_2^0 = ?$

Step 1. $T=175+273=448$ K; treating the % composition at equilibrium as proportional to concentration, $K(B\to A)=[A]/[B]=1.3/95.2=0.01366$, and $K(B\to C)=[C]/[B]=3.5/95.2=0.03676$.

Step 2. $\Delta G_1^0=-RT\ln K(B\to A)=-8.314\times448\times\ln(0.01366)=-3724.7\times(-4.293)\approx+15{,}994$ J $\approx+16.0$ kJ mol$^{-1}$.

Step 3. $\Delta G_2^0=-RT\ln K(B\to C)=-8.314\times448\times\ln(0.03676)=-3724.7\times(-3.303)\approx+12{,}302$ J $\approx+12.3$ kJ mol$^{-1}$. …

Read each equilibrium constant directly off the given percentage composition (ratio of the two species in that specific equilibriu …

• Using the wrong pair of percentages for each $K$ (the two equilibria, B$\rightleftharpoons$A and B$\rightleftharpoons$C, each use only …