Question

Calculate the equilibrium constatant $\left(K_C\right)$ at 1000 K . For the reaction $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$
Given that pressure of $C{O}_2$ is $2.4\times {10}^4\text{Pa}$.
Take R = 0.08

• A. 0.003
• B. 0.0056
• C. 0.03
• D. 0.28

Answer 1

The correct option is A 0.003

$CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$
1ng = 1 (Only gases are conbridered)
$\therefore K_P=P_{C{O}_2}=2.3\times {10}^4\text{Pa}$
$=\frac{2.4\times {10}^4}{{10}^5}\text{atm}(1Pa={10}^5\text{atm})$
$=0.24\text{atm}$
$K_P=R_C(RT{)}^{\Delta \left(g\right)}=K_c.(RT{)}^1$
$\Rightarrow K_C=\frac{K_P}{RT}=\frac{0.24\text{atm}}{0.08{\text{L.atm K}}^{-1}\text{mol}\times 1000K}$
= 0.003