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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.

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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.

△ 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.

Δ H

and

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for the system

H_{2} O (l) ⇌ H_{2} O (g)

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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 rates of forward and backward reactions will be the same. Predict the sign of free energy for this transformation above this temperature.

Q. For vaporization of water at

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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.

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Gibbs Free Energy & Spontaneity

Standard XII Chemistry

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△H and △S for the equilibrium reaction, H2O(l)⇌H2O(g) at 1 atmospheric pressure are 40.63 kJmol−1 and 108.8 JK−1mol−1 respectively. Calculate the temperature at which the reaction occur. Also, predict the sign of free energy for this transformation above this temperature.