For the reaction CaCO3(s)⇌CaO(s)+CO2(g), Kn=1.16atm at 1160K if 20g of CaCO3 in 10.0L vessel at 1160 K, is taken the fraction of CaCO3 dissociated is

For the reaction ${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g), K_{n} = 1.16 atm at 1160 K$ if $20 g of {CaCO}_{3} in 10.0 L$ vessel at 1160 K, is taken the fraction of ${CaCO}_{3}$ dissociated is

A
0.52
B
0.61
C
0.71
D
0.75

Solution:

$K_{p} = 1.16 a t m$ implies $p_{{CO}_{2}} = 1.16 atm$

The amount of ${CO}_{2}$ is $n = \frac{p V}{R T} = \frac{(1.16 atm) (10 L)}{(0.082 L atm K^{- 1} {mol}^{- 1}) (1160 K)} = \frac{1}{8.2} mol$

Mass of ${CaCO}_{3}$ dissociated, $m = n M = (\frac{1}{8.2} mol) (100 g {mol}^{- 1}) = 12.2 g$

Fraction of ${CaCO}_{3}$ dissociated $= \frac{12.2 g}{20 g} = 0.61$

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