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Q.

At 819°C and 1 atm, equilibrium is attained for the reaction, ${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$ in a closed vessel.Initially ‘X’ moles of ${CaCO}_{3}$ was present in the vessel. Equilibrium constant (Kc) for the reaction is

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a

$\frac{1}{89.6} M$

b

$\frac{1}{67.2} M$

c

$\frac{1}{44.8} M$

d

$\frac{1}{33.6} M$

answer is A.

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Detailed Solution

$\begin{array}{l} Equilibrium constant (K_{c}) = [{CO}_{2}] \\ {[{CaCO}_{3}]}_{solid} = {[CaO]}_{solid} = 1 \\ For an ideal gas, PV = nRT; \\ \frac{n}{V} = \frac{P}{RT} \\ \frac{n}{V} = [{CO}_{2}] \\ [{CO}_{2}] = \frac{P}{RT} \\ [{CO}_{2}] = \frac{1}{(0 .0821) (273 \times 4)} \\ [{CO}_{2}] = \frac{1}{22 .4 \times 4} M \\ K_{c} = \frac{1}{89 .6} M \end{array}$

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