$Δ H$ 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.

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Solution

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

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Similar questions

△ 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 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.

Q. For the reaction,

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

Calculate the temperature at which free energy change is zero. At a temperature lower than this, predict whether the forward of the reverse reaction will be forward. Give reason
(

Δ H = + 30.56 k J; Δ S == + 0.066 k J K^{- 1} {m o l}^{- 1}

)

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ΔH and ΔS for the reaction,Ag2O(s)⟶2Ag(s)+12O2(g) are 30.56kJmol−1 and 66.0JK−1mol−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.

We know thatΔG=ΔH−TΔSAt equilibrium ΔG=0so that 0=ΔH−TΔSor T=ΔHΔSGiven that, ΔH=30.56kJmol−1=30560Jmol−1ΔS=66.0JK−1mol−1T=3056066=463KAbove this temperature, ΔG will be negative and the process will be spontaneous in forward direction.