Q.54. Consider the following data for the reaction at 600 K. The (in kJ mol⁻¹) for the reaction is

Q: What is the correct answer for JEE Main Chemistry Q54?

The correct answer is B. Given reaction: $X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)$ Temperature $T = 600$ K From the table: For XY(g): $\Delta H^\circ = 42$ kJ/mol, $S^\circ = 200$ J/mol.K For $X_2(g)$: $\Delta H^\circ = 8$ kJ/mol, $S^\circ = 140$ J/mol.K For $Y_2(g)$: $\Delta H^\circ = 80$ kJ/mol, $S^\circ = 250$ J/mol.K Calculate $\Delta H^\circ_{reaction}$: $\Delta H^\circ_{reaction} = \sum (\Delta H^\circ_{products

Q: Q.54. Consider the following data for the reaction at 600 K. The (in kJ mol⁻¹) for the reaction is: [Figure: Table with A/H600K and S600K values for XY(g), X2(g), Y2(g)]

The correct answer is B. Given reaction: $X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)$ Temperature $T = 600$ K From the table: For XY(g): $\Delta H^\circ = 42$ kJ/mol, $S^\circ = 200$ J/mol.K For $X_2(g)$: $\Delta H^\circ = 8$ kJ/mol, $S^\circ = 140$ J/mol.K For $Y_2(g)$: $\Delta H^\circ = 80$ kJ/mol, $S^\circ = 250$ J/mol.K Calculate $\Delta H^\circ_{reaction}$: $\Delta H^\circ_{reaction} = \sum (\Delta H^\circ_{products

Q: What are the options for this JEE Main Chemistry question?

A. -21000 · B. -10 · C. -1000 · D. -9.012

Question

Q.54. Consider the following data for the reaction  at 600 K. The  (in kJ mol⁻¹) for the reaction is: [Figure: Table with A/H600K and S600K values for XY(g), X2(g), Y2(g)]

FirstInTest · Accepted Answer

Given reaction: $X_2(g) + Y_2(g) ightleftharpoons 2XY(g)$
Temperature $T = 600$ K

From the table:
For XY(g): $\Delta H^\circ = 42$ kJ/mol, $S^\circ = 200$ J/mol.K
For $X_2(g)$: $\Delta H^\circ = 8$ kJ/mol, $S^\circ = 140$ J/mol.K
For $Y_2(g)$: $\Delta H^\circ = 80$ kJ/mol, $S^\circ = 250$ J/mol.K

Calculate $\Delta H^\circ_{reaction}$:
$\Delta H^\circ_{reaction} = \sum (\Delta H^\circ_{products}) - \sum (\Delta H^\circ_{reactants})$
$\Delta H^\circ_{reaction} = [2 	imes \Delta H^\circ_{f}(XY)] - [\Delta H^\circ_{f}(X_2) + \Delta H^\circ_{f}(Y_2)]$
$\Delta H^\circ_{reaction} = [2 	imes 42] - [8 + 80] = 84 - 88 = -4$ kJ/mol.

Calculate $\Delta S^\circ_{reaction}$:
$\Delta S^\circ_{reaction} = \sum (S^\circ_{products}) - \sum (S^\circ_{reactants})$
$\Delta S^\circ_{reaction} = [2 	imes S^\circ(XY)] - [S^\circ(X_2) + S^\circ(Y_2)]$
$\Delta S^\circ_{reaction} = [2 	imes 200] - [140 + 250] = 400 - 390 = 10$ J/mol.K.
Convert to kJ/mol.K: $\Delta S^\circ_{reaction} = 10 	imes 10^{-3}$ kJ/mol.K = 0.010 kJ/mol.K.

Now, calculate $\Delta G^\circ_{reaction}$ using the Gibbs-Helmholtz equation:
$\Delta G^\circ_{reaction} = \Delta H^\circ_{reaction} - T \Delta S^\circ_{reaction}$
$\Delta G^\circ_{reaction} = -4 	ext{ kJ/mol} - (600 	ext{ K} 	imes 0.010 	ext{ kJ/mol.K})$
$\Delta G^\circ_{reaction} = -4 - 6 = -10$ kJ/mol.

The value of $\Delta G^\circ$ for the reaction is -10 kJ/mol.

Answer

-21000

Answer

-10

Answer

-1000

Answer

-9.012

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