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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

Δ H and Δ S...

Question

$Δ H$ and $Δ S$ 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 $66.00 J K^{- 1} m o l^{- 1}$ respectively. The temperature at which the free energy change for the reaction will be zero is :

393 K

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463 K

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413 K

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493 K

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Solution

The correct option is C $463 K$
$Δ G = Δ H - T Δ S$
$0 = Δ H - T Δ S$
$Δ H = T Δ S$
$∴ T = \frac{Δ H}{Δ S} = \frac{30560 J / m o l}{66.00 J / K / m o l} = 463 K$
Hence, the temperature at which the free energy change for the reaction will be zero is $463 K$ .

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Similar questions

Q. For reaction,

{A g}_{2} O (s) \to 2 A g (s) + (1 / 2) O_{2} (g)

, the value of

Δ H = 30.56 k J {m o l}^{- 1}

and

Δ S = 66 J K^{- 1} {m o l}^{- 1}

. Temperature at which free energy change for reaction will be zero, is:

D e l t a H

and

D e l t a 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 66.00

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

respectively, the temperature at which the free energy change for the reaction will be zero is:

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.

$△ H$ 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 (s) ⇌ 2 A g (s) + \frac{1}{2} O_{2} (g)

Δ H, Δ S, Δ T

are

40.63 k J {m o l}^{- 1}, 1.808 J K^{- 1} {m o l}^{- 1}

and

373.4 K

respectively. Free energy change

Δ G

of the reaction will be:

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ΔH and ΔS for the reaction, Ag2O(s)→2Ag(s)+12O2(g), are 30.56 kJ mol−1 and 66.00 JK−1 mol−1 respectively. The temperature at which the free energy change for the reaction will be zero is :

The correct option is C 463 KΔG=ΔH−TΔS0=ΔH−TΔSΔH=TΔS∴T=ΔHΔS=30560J/mol66.00J/K/mol=463KHence, the temperature at which the free energy change for the reaction will be zero is 463 K.

$Δ H$ and $Δ S$ 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 $66.00 J K^{- 1} m o l^{- 1}$ respectively. The temperature at which the free energy change for the reaction will be zero is :

The correct option is C $463 K$
$Δ G = Δ H - T Δ S$
$0 = Δ H - T Δ S$
$Δ H = T Δ S$
$∴ T = \frac{Δ H}{Δ S} = \frac{30560 J / m o l}{66.00 J / K / m o l} = 463 K$
Hence, the temperature at which the free energy change for the reaction will be zero is $463 K$ .