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Calculate δGθ For the Reaction, and Predict Whether the Reaction May Occur Spontaneously. - CBSE (Science) Class 11 - Chemistry

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Question

For the reaction

2A(g) + B(g) → 2D(g)

ΔUθ = –10.5 kJ and ΔSθ= –44.1 JK–1.

Calculate ΔGθ for the reaction, and predict whether the reaction may occur spontaneously.

Solution 1

For the given reaction,

2 A(g) + B(g) → 2D(g)

Δng = 2 – (3)

= –1 mole

Substituting the value of ΔUθ in the expression of ΔH:

ΔHθ = ΔUθ + ΔngRT

= (–10.5 kJ) – (–1) (8.314 × 10–3 kJ K–1 mol–1) (298 K)

= –10.5 kJ – 2.48 kJ

ΔHθ = –12.98 kJ

Substituting the values of ΔHθ and ΔSθ in the expression of ΔGθ:

ΔGθ = ΔHθ – TΔSθ

= –12.98 kJ – (298 K) (–44.1 J K–1)

= –12.98 kJ + 13.14 kJ

ΔGθ = + 0.16 kJ

Since ΔGθ for the reaction is positive, the reaction will not occur spontaneously.

Solution 2

`triangleH^(Theta) = triangleU^(Theta) + triangle^"ng" RT`

`triangleU^(Theta) = -10.5 kJ; triangle^"ng" = 2  - 3` =  -1 mol

`R = 8.314 xx 10^(-3) kJ K^(-1) mol^(-1)`; T = 298 K

`:. triangle H^(Theta) = (-10.5 kJ) + [(- 1 mol) xx (8.314 xx 10^(-3) kJ K^(-1) mol^(-1))xx(298K)]`

= -10.5 kJ - 2.478 kJ = -12.978 kJ

According to Gibbs Helmhoitz equation:

`triangle G^(Theta) = triangleH^(Theta) - T triangle S^(Theta)`

`triangle G^(Theta) = (-12.978 kJ) - (298 K) xx (-0.0441 kJ K^(-1))`

= - 12.978 + 13.112 = - 12.978 + 13.142 = 0.164 kJ

Since `triangleG^(Theta)` is postive the reaction is non-spontaneous in nature

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Solution Calculate δGθ For the Reaction, and Predict Whether the Reaction May Occur Spontaneously. Concept: Spontaneity - Is Decrease in Enthalpy a Criterion for Spontaneity.
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