JEE Advanced levelSingle correct optionDifficulty 3 of 3td215
At 310K a reaction AB has ΔG∘=+30.0kJmol−1 and is coupled with CD with ΔG∘=−50.0kJmol−1. Taking R=8.314Jmol−1K−1, the equilibrium constant of the coupled reaction A+CB+D is nearest to
A4.3×10−4
B2.7×108
C3.2×103
D2.3×103
Answer: D
For coupled reactions ΔG∘ values add: 30.0+(−50.0)=−20.0kJmol−1. Then K=exp(20000/(8.314×310))=exp(7.760)=2.3×103. Using 298K or the wrong sign gives the other values.