For the reaction $\mathrm{CH}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{C}_2…

For the reaction $\mathrm{CH}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{C}_2 \mathrm{H}_{6(\mathrm{~g})}$ $\mathrm{K}_{\mathrm{p}}=3.356 \times 10^{17}$, calculate $\Delta \mathrm{G}^{\circ}$ for the reaction at 298 K .
  1. $-90 \mathrm{~kJ~mol}^{-1}$
  2. $-100 \mathrm{~kJ~mol}^{-1}$
  3. $235.6 \mathrm{~kJ~mol}^{-1}$
  4. $33.56 \mathrm{~kJ~mol}^{-1}$

Solution

\(\begin{aligned} & \Delta G^0=-2.303 R T \log _{10} K_p \\ & R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, T=298 \mathrm{~K} \\ & K_P=3.356 \times 10^{17} \\ & \Delta G^0=-2.303 \times 8.314 \times 298 \times \log _{10}\left(3.356 \times 10^{17}\right) \\ & =-2.303 \times 8.314 \mathrm{Jmol}^{-1} \times 298 \times 17.526 \\ & =-100,000 \mathrm{Jmol}^{-1} \\ & =-100 \mathrm{kJmol}^{-1}\end{aligned}\)

Asked in: MHT CET 2024 (10 May Shift 2)

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