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Derivation of Kp and Kc

The standard ...

Question

The standard enthalpy and entropy changes for the reaction in equilibrium for the forward direction are given below:
$C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)$
$△ H_{300 K}^{o} = - 41.16 k J m o l^{- 1}$
$△ S_{300 K}^{o} = - 4.24 \times 10^{- 2} k J m o l^{- 1}$
Calculate $K_{p}$ (at 300 K.)

A

4.93 \times 10^{4}

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B

8.93 \times 10^{- 4}

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C

8.93 \times 10^{4}

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D

None of the above

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Solution

The correct option is C $8.93 \times 10^{4}$
At 300 K: $△ G^{o} = △ H^{o} - T △ S^{o}$
$= - 41.16 - 300 \times (- 4.24 \times 10^{- 2})$
$= - 28.44 k J$
we know
$△ G^{o} = - 2.303 R T l o g K_{P}$
$- 28.44 = - 2.303 \times 8.314 \times 10^{- 3} \times 300 l o g K_{P}$
$K_{P} = 8.93 \times 10^{4}$

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Derivation of Kp and Kc

Standard XIII Chemistry

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