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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

H and S for t...

Question

$△ H$ and $△ S$ for the equilibrium reaction, $H_{2} O (l) ⇌ H_{2} O (g)$ at 1 atmospheric pressure are $40.63 k J m o l^{- 1}$ and $108.8 J K^{- 1} m o l^{- 1}$ respectively. Calculate the temperature at which the reaction occur.
Also, predict the sign of free energy for this transformation above this temperature.

373.4 K

, negative

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373.4 K

, positive

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463.4 K

, negative

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463.4 K

, positive

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Solution

The correct option is A $373.4 K$ , negative
Given:
$△ H = 40.63 k J m o l^{- 1}$
$△ S = 108.8 J K^{- 1} m o l^{- 1} = 0.1088 k J K^{- 1} m o l^{- 1}$

$△ G = 0$ (as the system is in equilibrium)
Applying Gibb's free energy equation:
$△ G = △ H - T △ S$
$0 = 40.63 - T \times 0.1088$
$T = \frac{40.63}{0.1088} = 373.4 K$

The sign of $△ G$ above this temperature will become negative as magnitude of $(T Δ S > Δ H)$ and the reaction will be spontaneous.

Suggest Corrections

Similar questions

△ H

and

△ S

for the equilibrium reaction,

H_{2} O (l) ⇌ H_{2} O (g)

at 1 atmospheric pressure are

40.63 k J m o l^{- 1}

and

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

respectively. Calculate the temperature at which the reaction occur.
Also, predict the sign of free energy for this transformation above this temperature.

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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.