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Reaction Quotient

If standard e...

Question

If standard entropies of $N_{2}, H_{2}$ and $N H_{3}$ are 60, 40, 50 $J / (K . m o l)$ respectively and enthalpy of formation of $N H_{3}$ is $- 30 k J . m o l^{- 1}$ , then the temperature at which reaction for formation of ammonia will be at equilibrium is:

A

227^{o} C

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B

477^{o} C

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C

750^{o} C

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D

900^{o} C

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Solution

The correct option is D $477^{o} C$
$\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⟶ N H_{3}$

$Δ S_{r e a c t i o n} = 50 - [\frac{3}{2} \times 40 + \frac{1}{2} \times 60] = - 40 J m o l^{- 1}$

$Δ G = Δ H - T Δ S$

At equilibrium ; $Δ G = 0$

So, $Δ H = T Δ S$

$\Rightarrow 30 \times 10^{3} = T \times 40$

$\Rightarrow T = 750 K$ or $477^{o} C$

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