JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Consider the following data for the reaction

$X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)$

at $600\text{ K}$ . The $\Delta_r G^\circ$ (in kJ mol $^{-1}$ ) for the reaction is :

Compound	$\Delta_f H^\circ_{600K}$ (kJ mol $^{-1}$ )	$S^\circ_{600K}$ (J mol $^{-1}$ K $^{-1}$ )
$XY(g)$	$42$	$200$
$X_2(g)$	$8$	$140$
$Y_2(g)$	$80$	$250$

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$-21000$
B
$-10$
C
$-1000$
D
$-9.012$

Did you get this right?

Solution

The given reaction is $X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)$ .

The standard enthalpy of reaction, $\Delta_r H^\circ$ , is calculated as:

$\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$

$\Delta_r H^\circ = 2 \Delta_f H^\circ(XY) - [\Delta_f H^\circ(X_2) + \Delta_f H^\circ(Y_2)]$

$\Delta_r H^\circ = 2(42) - (8 + 80) = 84 - 88 = -4 \text{ kJ mol}^{-1}$

The standard entropy of reaction, $\Delta_r S^\circ$ , is calculated as:

$\Delta_r S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})$

$\Delta_r S^\circ = 2 S^\circ(XY) - [S^\circ(X_2) + S^\circ(Y_2)]$

$\Delta_r S^\circ = 2(200) - (140 + 250) = 400 - 390 = 10 \text{ J K}^{-1} \text{ mol}^{-1}$

Converting $\Delta_r S^\circ$ to $\text{kJ K}^{-1} \text{ mol}^{-1}$ :

$\Delta_r S^\circ = \dfrac{10}{1000} = 0.01 \text{ kJ K}^{-1} \text{ mol}^{-1}$

Now, using the Gibbs free energy equation:

$\Delta_r G^\circ = \Delta_r H^\circ - T \Delta_r S^\circ$

$\Delta_r G^\circ = -4 - 600 \times 0.01$

$\Delta_r G^\circ = -4 - 6 = -10 \text{ kJ mol}^{-1}$

Answer: $-10$

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Compound

\Delta_f H^\circ_{600K}

(kJ mol

^{-1}

)

S^\circ_{600K}

(J mol

^{-1}

^{-1}

)

XY(g)

42

200

X_2(g)

8

140

Y_2(g)

80

250