Calculate G^{} for reaction, {CH}₄(g)}+{H}₂(g)} arrow C₂ {H}₆(g)}≤ft(K_{p}=2 × 10¹7}, R=8.314 {JK}^{-1} {~mol}^{-1})

Correct Answer is : -98.716 kJ mol-1

Correct Answer is : -98.716 kJ mol-1

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MHTCET 20 Apr 2026 Shift 1 Paper

Question : 58 of 150

Marks: +1, -0

Calculate

\triangle G^{\circ}

\mathrm{CH}_{4(g)}+\mathrm{H}_{2(g)} \rightarrow C_{2} \mathrm{H}_{6(g)}\left(K_{p}=2 \times 10^{17}, R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)

Solution:

Concept:

The standard Gibbs energy change is related to the equilibrium constant by the equation

\Delta G^{\circ}=-RT\ln K_p

Explanation:

At standard temperature,

T=298\ \mathrm{K}

Given

R=8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}

and

K_p=2\times10^{17}

\Delta G^{\circ}=-(8.314)(298)\ln(2\times10^{17})

Now compute the natural logarithm:

\ln(2\times10^{17})=\ln 2+17\ln 10

=0.693+17(2.303)=39.844

Substitute this value into the equation:

\Delta G^{\circ}=-(8.314)(298)(39.844)

\Delta G^{\circ}\approx -98716\ \mathrm{J\ mol^{-1}}

Convert to kilojoules:

\Delta G^{\circ}=-98.716\ \mathrm{kJ\ mol^{-1}}

The negative value indicates the process is spontaneous under standard conditions.

Answer:

Option B:

-98.716\ \mathrm{kJ\ mol^{-1}}

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