Use the following data One mole each of A 2 g and B 2 g... | null (Chemistry)

JEE Mains 2026 21st January Shift 1

Chemistry

Use the following data : TABLE One mole each of

\mathrm{A}_{2}(\mathrm{~g})

and

\mathrm{B}_{2}(\mathrm{~g})

are taken in a 1L closed flask and allowed to establish the equilibrium at 500 K .

\\ \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})\\

The value of

\mathrm{x}\left(\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}\right)

\_\_\_\_

(Nearest integer)

\\

(Given:

\log \mathrm{K}=2.2 \mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}

)

\mathrm{A}_{2}+\mathrm{B}_{2} \xrightarrow{500 \mathrm{~K}} 2 \mathrm{AB} \log \mathrm{K}=2.2\\

\Delta \mathrm{H}^{\circ}=(2 \times 32)-(6+\mathrm{x})=(58-\mathrm{x}) \mathrm{kJ}\\

\Delta \mathrm{S}^{\circ}=(2 \times 222)-(146+280)=18

Joule

\\ \Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}\\

\Delta \mathrm{G}^{\circ}=-\frac{8.314 \times 500 \times 2.2 \times 2.303}{1000}\\

\Delta \mathrm{G}^{\circ}=-21.06\\

\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}=-21.06\\

58-\mathrm{x}-500\left(\frac{18}{1000}\right)=-21.06\\

\mathrm{x}=70.06 \mathrm{KJ} / \mathrm{mol}

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