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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
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B
0.145 atm
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C
2.146 atm
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D
3.145 atm
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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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