Question

# When $$20\ g$$ of $$CaCO_{3}$$ were put in a $$10\ litre$$ flask and heated to $$800^{\circ} C$$ , $$35\%$$ of $$CaCO_{3}$$ remained unreacted at equilibrium. $$K_{P}$$ for decomposition of  $$CaCO_{3}$$ will be

A
1.145 atm
B
0.145 atm
C
2.146 atm
D
3.145 atm

Solution

## The correct option is D $$1.145\ atm$$The equilibrium reaction is $$CaCO_{3}(s) \Leftrightarrow CaO(s) + CO_{2}(g)$$.The following table lists the quantities of various species.$$CaCO_3$$$$CaO$$ $$CO_2$$Initial mass (g)$$20$$$$0$$         $$0$$Initial moles$$0.2$$$$0$$ $$0$$Moles at equilibrium $$0.35 \times 0.2$$     $$0.13$$$$0.13$$The ideal gas equation is $$PV=nRT$$.The partial pressure of $$CO_2=P=\dfrac {nRT} {V}= \dfrac {0.13 \times 0.0821 \times 1073} {10} =1.145$$ atm.The expression for the equilibrium constant $$(K_P )$$ $$= P_{CO_2}=1.145$$ atm.Chemistry

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