$△ H$ and $△ S$ for the reaction: $A g_{2} O (s) \to 2 A g (s) + \frac{1}{2} O_{2} (g)$ are 30.56 kJ $m o l^{- 1}$ and 66.0 J $K^{- 1} m o l^{- 1}$ respectively. Calculate the temperature at which free energy change for the reaction will be zero. Predict whether the forward reaction will be favored above or below this temperature.

298 K, above

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300 K, below

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373 K, above

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463 K, above

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Solution

The correct option is D
463 K, above

We know that, $△ G = △ H - T △ S$
At equilibrium, $△ G$ = 0
So that 0 = $△ H - T △ S$
or $T = \frac{△ H}{△ S}$
given that $△ H$ - 30.56 kJ $m o l^{- 1}$ = 30560 J $m o l^{- 1}$
$△ S$ = 66.0 J $K^{- 1} m o l^{- 1}$
T = $\frac{30560}{66}$ = 463 K
Above this temperature, $△ G$ will be negative and the process will be spontaneous in forward direction.

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Δ H

and

Δ S

for the reaction,

{A g}_{2} O (s) ⟶ 2 A g (s) + \frac{1}{2} O_{2} (g)

are

30.56 k J {m o l}^{- 1}

and

66.0 J K^{- 1} {m o l}^{- 1}

respectively. Calculate the temperature at which free energy change for the reaction will be zero. Predict whether the forward reaction will be forward above or below this temperature.

△ H a n d △ S

for the reaction,

A g_{2} O (s) \to 2 A g (s) + \frac{1}{2} O_{2} (g)

are

30.56 k J m o l^{- 1}

and

66.0 J K^{- 1}

respectively. Calculate the temperature at which free energy change for the reaction will be zero. Predict whether the forward reaction will be favoured above or below this temperature (in K)

$△ H$ and $△ S$ for the reaction: $A g_{2} O (s) \to 2 A g (s) + \frac{1}{2} O_{2} (g)$ are 30.56 kJ $m o l^{- 1}$ and 66.0 J $K^{- 1} m o l^{- 1}$ respectively. Calculate the temperature at which free energy change for the reaction will be zero. Predict whether the forward reaction will be favored above or below this temperature.