For a given reaction:
$x A + y B \to m C + n D$
$△ H = - 30 k J m o l^{- 1}, △ S = - 100 J K^{- 1} m o l^{- 1}$
At what temperature the reaction is at equilibrium?

50^{o} C

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250^{o} C

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100^{o} C

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27^{o} C

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Solution

The correct option is D $27^{o} C$
Gibb's free energy of a reaction is given by:
$△ G = △ H - T △ S$
we know at equilibrium, $△ G = 0$
$∴ T = \frac{△ H}{△ S} = \frac{- 30 \times 1000}{- 100} = 300 K, o r, 27^{o} C$
hence, the reaction is at equilibrium at $27^{o} C$

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x A + y B \to m C + n D

△ H = - 30 k J m o l^{- 1}, △ S = - 100 J K^{- 1} m o l^{- 1}

x A + y B \to m C + n D

△ H = - 30 k J m o l^{- 1}, △ S = - 100 J K^{- 1} m o l^{- 1}

P + Q \to R + S .

△ H^{o} i s - 30 k J m o l^{- 1} a n d △ S i s - 100 J K^{- 1} m o l^{- 1} .

P + Q \to R + S

Δ H^{0}

m o l^{- 1}

Δ S

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

△ H = + 29 k J m o l^{- 1}; △ S = - 35 J K^{- 1} m o l^{- 1}

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