Search any question & find its solution
Question:
Answered & Verified by Expert
If equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} \mathrm{C}$, standard free energy change of the process is
$$
\left(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 0.0038=-2.42\right)
$$
Options:
$$
\left(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 0.0038=-2.42\right)
$$
Solution:
1833 Upvotes
Verified Answer
The correct answer is:
$13.8 \mathrm{~kJ} \mathrm{~mol}^1$
$$
\begin{aligned}
\Delta G^{\circ} & =-R T \ln K \\
\Delta G^{\circ} & =-2.303 R T \log K \\
\Delta G^{\circ} & =-2.303 \times 8.314 \mathrm{~mol}^{-1} \times 298 \log 3.8 \times 10^{-3} \\
\Delta G^{\circ} & =13.8 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} \\
\Delta G^{\circ} & =13.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
\begin{aligned}
\Delta G^{\circ} & =-R T \ln K \\
\Delta G^{\circ} & =-2.303 R T \log K \\
\Delta G^{\circ} & =-2.303 \times 8.314 \mathrm{~mol}^{-1} \times 298 \log 3.8 \times 10^{-3} \\
\Delta G^{\circ} & =13.8 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} \\
\Delta G^{\circ} & =13.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.