Ncert Exercises & Solutions

Q. 19. For the reaction

$2 A_{(g)} + B_{(g)} \to 2 D_{(g)} ∆ U^{θ} = - 10.5 k J a n d ∆ S^{θ} = - 44.1 J K^{- 1}$

Calculate $∆ G^{θ}$ for the reaction, and predict whether the reaction may occur spontaneously.

NEETprep Answer:

For the given reaction,

2 A_{(g)} + B_{(g)} \to 2 D_{(g)} ∆ n_{g} = 2 - (3) = - 1 m o l e

Substituting the value of

∆ U^{θ}

in the expression of

∆ H

∆ H^{θ}

∆ U^{θ}

∆ n_{g} R T

= (- 10.5 k J) - (- 1) (8.314 \times 10^{- 3} k J K^{- 1} m o l^{- 1}) (298 K) = - 10.5 k J - 2.48 k J ∆ H^{θ} = - 12.98 k J

Substituting the values of $∆ H^{θ}$ and $∆ S^{θ}$ in the expression of $∆ G^{θ}$ :
$∆ G^{θ}$ = $∆ H^{θ}$ -T $∆ S^{θ}$

$= - 12.98 k J - (- 298 K) (- 44.1 J K^{- 1}) = - 12.98 k J + 13.14 k J ∆ G^{θ} = + 0.16 k J$

Since $∆ G^{θ}$ for the reaction is positive, the reaction will not occur spontaneously.