$△ H a n d △ S$ 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)

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Solution

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 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}

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△ H a n d △ S

for the reaction,

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

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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 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

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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)

$△ H$ 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.

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△H and △S for the reaction, Ag2O(s)→2Ag(s)+12O2(g) are 30.56kJ mol−1 and 66.0J 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)

We know that, △G=△H−T △S At equilibrium, △G=0 so that 0=△H−T △S or T=△H△S Given that, △H=30.56 kJ mol−1 =30560 J mol−1 △S=66.0 J K−1mol−1 T=3056066=463K Above this temperature, △G will be negative and the process will be spontaneous in forward direction.