Equilibrium - Result Question 16
####17. Value of $K_P$ in the reaction
$MgCO_3(s) \to MgO(s)+CO_2(g)$ is
[2000]
(a) $K_P=P _{CO_2}$
(b) $K_P=P _{CO_2} \times \frac{P _{CO_2} \times P _{MgO}}{P _{MgCO_3}}$
(c) $K_P=\frac{P _{CO_2} \times P _{MgO}}{P _{MgCO_3}}$
(d) $K_P=\frac{P _{MgCO_3}}{P _{CO_2} \times P _{MgO}}$
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Solution:
- (a) $MgCO_3$ (s) $arrow MgO$ (s) $+CO_2$ (g)
$MgO & MgCO_3$ are solid and they do not exert any pressure and hence only pressure exerted is by $CO_2$. Therefore $K_P=P _{CO 2}$
