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 $^{-1}$ )
$XY(g)$ $42$ $200$
$X_2(g)$ $8$ $140$
$Y_2(g)$ $80$ $250$

Question

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 $^{-1}$ )
$XY(g)$	$42$	$200$
$X_2(g)$	$8$	$140$
$Y_2(g)$	$80$	$250$

Quantrex Academy · Accepted Answer

Correct answer: B. -10

The given reaction is X_2(g) + Y_2(g) leftharpoons 2XY(g) . The standard enthalpy of reaction, Δ_r H^ , is calculated as: Δ_r H^ = ∑ Δ_f H^ (products) - ∑ Δ_f H^ (reactants) Δ_r H^ = 2 Δ_f H^ (XY) - [Δ_f H^ (X_2) + Δ_f H^ (Y_2)] Δ_r H^ = 2(42) - (8 + 80) = 84 - 88 = -4 kJ mol^ -1 The standard entropy of reaction, Δ_r S^ , is calculated as: Δ_r S^ = ∑ S^ (products) - ∑ S^ (reactants) Δ_r S^ = 2 S^ (XY) - [S^ (X_2) + S^ (Y_2)] Δ_r S^ = 2(200) - (140 + 250) = 400 - 390 = 10 J K^ -1 mol^ -1 Converting Δ_r S^ to kJ K^ -1 mol^ -1 : Δ_r S^ = (10)/(1000) = 0.01 kJ K^ -1 mol^ -1 Now, using the Gibbs free energy equation: Δ_r G^ = Δ_r H^ - T Δ_r S^ Δ_r G^ = -4 - 600 × 0.01 Δ_r G^ = -4 - 6 = -10 kJ mol^ -1 Answer: -10

Consider the following data for the reaction X_2(g) + Y_2(g)...

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