- H and S for the reaction- Ag 2 O s - 2 Ag s -1-2 O 2 g are 30-56 kJ mol -1 and 66-0 J mol -1 K -1 respectively- Calculate the temperature at which free energy change for the reaction will be zero- Predict whether the reaction will be sponteneous above or below this temperature in K A- 563 K- reaction will be sponteneous above this temperature-B- 463 K- reaction will be sponteneous above this temperature-C- 563 K- reaction will be sponteneous below this temperature-D- 463 K- reaction will be sponteneous below this temperature-

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Byju's Answer

Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

Δ H and S for...

Question

$△ H a n d △ S$ 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 $66.0 J m o l^{- 1} K^{- 1}$ respectively. Calculate the temperature at which free energy change for the reaction will be zero. Predict whether the reaction will be sponteneous above or below this temperature (in K)

563 K

, reaction will be sponteneous above this temperature.

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463 K

, reaction will be sponteneous above this temperature.

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563 K

, reaction will be sponteneous below this temperature.

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463 K

, reaction will be sponteneous below this temperature.

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Solution

The correct option is B $463 K$ , reaction will be sponteneous above this temperature.
We know that,
$△ G = △ H - T △ S$
At equilibrium, $△ G = 0$
so that $0 = △ H - T △ S$
or $T = \frac{△ H}{△ S}$
Given that, $△ H = 30.56 k J m o l^{- 1}$
$= 30560 J m o l^{- 1}$
$△ S = 66.0 J K^{- 1} m o l^{- 1}$
$T = \frac{30560}{66} = 463 K$
Above this temperature, $△ G$ will be negative and the reaction will be spontaneous.

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△ H a n d △ S

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

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$△ H$ 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.