Question

For the reaction $1g$ mole of $Ca{CO}_3$ is enclosed in $5L$ container
$Ca{CO}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+{CO}_2\left(g\right);K_p=1.16atm$ at $1073K$ then percentage dissociation of $Ca{CO}_3$ is:

• A. $65$%
• B. $100$%
• C. $6.5$%
• D. Zero

Answer 1

The correct option is C $6.5$%

For the reaction $1g$ mole of $Ca{CO}_3$ is enclosed in $5L$ container
$Ca{CO}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+{CO}_2\left(g\right);K_p=1.16atm$ at $1073K$ then percentage dissociation of $Ca{CO}_3$ is $6.5$%
$K_p=P_{CO2}=1.16$ atm
The number of moles of $C{O}_2=$ $n=\frac{PV}{RT}$
$n=\frac{\text{1.16 atm}\times \text{5 L}}{\text{0.08206 L atm / mol /K}\times \text{1073 K}}$
$n=\text{0.065 mol}$
0.065 moles of $C{O}_2$ are formed when 0.065 moles of
$CaC{O}_3$ decompose.
The percentage dissociation of $Ca{CO}_3$ is $\frac{0.065}{1}\times 100=6.5$ %