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Q.

When 20 g of ${CaCO}_{3}$ were put in 10-liter flask and heated to 800°C,35% of ${CaCO}_{3}$ remained unreacted at equilibrium. $K_{P}$ for decomposition of ${CaCO}_{3} is$

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a

0.145atm

b

2.146atm

c

1.145 atm

d

3.145 atm

answer is A.

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Detailed Solution

${CaCO}_{3 (s)} ⇌ {CaO}_{(s)} + {CO}_{2 (g)}$

20 g (0.2 moles)

$0.2 \times \frac{65}{100} (65 % reacted)$

$K_{C} [{CO}_{2}] = \frac{0.13}{10} = 0.013$

$K_{p} = K_{C} {(RT)}^{1} = 0.013 \times 0.0821 \times 1073$

$K_{P} = 1.145 atm$

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