JEE Advanced levelSingle correct optionDifficulty 3 of 3ce155
Take R=8.314Jmol−1K−1. A reaction has ΔH∘=−92.0kJmol−1 and ΔS∘=−199Jmol−1K−1, both assumed independent of temperature. The value of log10K at 500K is closest to
A−0.78
B0.78
C3.42
D20.0
Answer: A
ΔG∘(500K)=ΔH∘−TΔS∘=−92000−500(−199)=+7500Jmol−1. log10K=−ΔG∘/(2.303RT)=−7500/(2.303×8.314×500)=−0.78. Using ΔG∘ at 298K gives 3.42, and a wrong sign on the TΔS∘ term gives 20.0.