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Question

The value of $$K_c$$ for the following equilibrium is $$CaCO_{3(s)} \rightleftharpoons CaO_(s) + CO_{2(g)}$$.
Given $$K_p =167$$ bar at $$1073$$K.


A
1.896molL1
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B
4.38×104molL1
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C
6.4×104molL1
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D
6.626molL1
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Solution

The correct option is A $$1.896 mol L^{-1}$$
$$ CaCO_3(s) \rightleftharpoons CaO(s) +CO_2(g)$$
$$\Delta n = (0+1)-0 =1$$
$$ K_c = \dfrac {K_p}{(RT)^{\Delta n}}$$
$$ K_c = \dfrac {167bar} {(0.0821Lbar/K) \times (1073K)}= 1.896 mol/L$$

Chemistry

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