JEE Main 2015 · 4 Apr · Shift 1 · Chemistry Q35

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ChemistrySingle correct+4 / -1Question 35 of 90

The following reaction is performed at 298 K.
$\ce{2NO(g) + O2(g) <=> 2NO2(g)}$
The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of $\ce{NO2(g)}$ at 298 K? ( $K_p=1.6\times10^{12}$ )

R(298)\ln(1.6\times10^{12})-86600

86600+R(298)\ln(1.6\times10^{12})

86600-\dfrac{\ln(1.6\times10^{12})}{R(298)}

0.5[2\times86{,}600-R(298)\ln(1.6\times10^{12})]

Answer and worked solution

Official answer: D

$\Delta G^\circ_r=2\Delta_fG(\ce{NO2})-2\Delta_fG(\ce{NO})=-RT\ln K_p$ , so
$\Delta_fG(\ce{NO2})=\frac12\left[2\times86600-R(298)\ln(1.6\times10^{12})\right].$

Topics:Equilibrium and Gibbs energy

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JEE Main 2015 · 4 Apr · Shift 1, Chemistry question 35