If equilibrium constant of a process is 3.8 \times 10^{-3} at 25^\circ{C}standard free energy change of the process is( {R}=8.314 {J} {mol}^{-1} {K}^{-1}, \log 0.0038=-2.42 )

Correct Answer is : 13.8kJmol−1

TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Chemistry

Question : 132 of 160

Marks: +1, -0

If equilibrium constant of a process is 3.8×10−3 at 25°Cstandard free energy change of the process is(R=8.314Jmol−1K−1, log‌0.0038=−2.42)

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