Question

$K_c$ for, $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$ is 13.7 mol litre${}^{-1}$ at $546K$. If the pressure developed in 10 litre box in equilibrium if reactant concentration is 1.0 mole is $P$. The value of $100P$ is_________.

Answer 1

$PC{l}_5\leftrightharpoons PC{l}_3+C{l}_2$

Initial conc.(M) $0.1$ $0$ $0$

At eq. $0.1-x$ $x$ $x$

The expression for the equilibrium constant is:

$K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}=\frac{x\times x}{0.1-x}=13.7$

Hence, $x=0.0993$

Total number of moles $=n=10(0.1+x)=10(0.1+0.0993)=1.993$

$P=\frac{nRT}{V}=\frac{1.993\times 0.0821\times 546}{10}$

$P=8.93$

$100P=893$