1) If equilibrium constant of a process is 3.8×10−3 at 250C , standard free energy change of the process is (R=8.314Jmol−1K−1,log0.0038=−2.42) A) 5.7kJmol−1 B) 9.9kJmol−1 C) 13.8kJmol−1 D) 15.6kJmol−1 Answer: Option CExplanation:△G0=−RTlnK △G0=−2.303RTlogK △G0=−2.303×8.314mol−1×298log3.8×10−3 △G0=13.8×103Jmol−1 △G0=13.8kJmol−1