TS EAMCET · Chemistry · Thermodynamics (C)
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 \[ \left(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 0.0038=-2.42\right) \]
- A \(5.7 \mathrm{~kJ} \mathrm{~mol}^1\)
- B \(9.9 \mathrm{~kJ} \mathrm{~mol}^1\)
- C \(13.8 \mathrm{~kJ} \mathrm{~mol}^1\)
- D \(15.6 \mathrm{~kJ} \mathrm{~mol}^1\)
Answer & Solution
Correct Answer
(C) \(13.8 \mathrm{~kJ} \mathrm{~mol}^1\)
Step-by-step Solution
Detailed explanation
\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} &…
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