Chemistry | TS EAMCET 2018 | Discussion Page

If equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{0} C$ , standard free energy change of the process is

$(R= 8.314 J mol^{-1} K^{-1}, \log 0.0038=-2.42$)

A) $5.7 kJ mol^{-1}$

B) $9.9 kJ mol^{-1}$

C) $13.8 kJ mol^{-1}$

D) $15.6 kJ mol^{-1}$

Option C

$\triangle G^{0}=-RT ln K$

$\triangle G^{0}=-2.303 RT \log K$

$\triangle G^{0}=-2.303 \times 8.314 mol^{-1} \times 298 \log 3.8 \times 10^{-3}$

$\triangle G^{0}=13.8 \times 10^{3} J mol ^{-1}$

$\triangle G^{0}=13.8 kJmol^{-1}$

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