Question

$50gCaC{O}_3$ are heated to $1073K$ in a $5L$ vessel. What percentage of $CaC{O}_3$ would decompose at equilibrium? The value of $K_p$ for the reaction.
$CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$ is $1.15atm$ at $1073K$.

Answer 1

Amount of $CaC{O}_3=50g$

Molar mass of $CaC{O}_3=40+12\times 3\times 16=100gm$

number of mole of $CaC{O}_3=\frac{50}{100}=0.5mole$

$CaC{O}_3\rightleftharpoons CaO+C{O}_2\left(g\right)$

$k_p=P_{C{O}_2}=1.15atm$

volume = $52$ temp = $1073k$

$n=\frac{P_{C{O}_2}v}{RT}=\frac{1.15\times 5}{0.0821\times 1073}=0.065$

$\%$ of $CaC{O}_3$ decompose = $\frac{0.065}{5}\times 100$

$=1.3\%$