For the reaction

2A(g) + B(g)→ 2D(g)

\( \Delta U^{\Theta } = -10.5 kJ \;and \;\Delta S^{\Theta } = -44.1JK^{-1}\)

Calculate \( \Delta G^{\Theta }\) for the reaction, and predict whether the reaction may occur spontaneously.

Asked by Abhisek | 1 year ago |  166

1 Answer

Solution :-

2A(g) + B(g) → 2D(g)

\( \Delta n_{g} = 2 – 3\)

= -1 mole

Putting value of \( \Delta U^{\Theta }\) in expression of \( \Delta H\):

\( \Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RT\)

= (-10.5KJ) – (-1)

\( ( 8.314\times 10^{-3}kJK^{-1}mol^{-1}\)(298K)

= -10.5kJ -2.48kJ

\( \Delta H^{\Theta }\) = -12.98kJ

Putting value of \( \Delta S^{\Theta }\) and\( \Delta H^{\Theta }\) in expression of\( \Delta G^{\Theta }\):

\( \Delta G^{\Theta } = \Delta H^{\Theta } – T\Delta S^{\Theta }\)

= -12.98kJ –(298K)\( (-44.1JK^{-1})\)

= -12.98kJ +13.14kJ

\( \Delta G^{\Theta }\) = 0.16kJ

As, \( \Delta G^{\Theta }\) is positive, the reaction won’t occur spontaneously.

Answered by Pragya Singh | 1 year ago

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