Question

At ${248}^o$C, the $K_p$ for the reaction, $SbC{l}_5\left(g\right)\rightleftharpoons SbC{l}_3\left(g\right)+C{l}_2\left(g\right)$ is $1.07$ atm at a total pressure of $1$ atm. Calculate the degree of dissociation of $SbC{l}_5$.

• A. $0.516$
• B. $0.718$
• C. $0.321$
• D. None of these

Answer 1

The correct option is B $0.718$

${SbC{l}_5}_{\left(g\right)}\rightleftharpoons {SbC{l}_3}_{\left(g\right)}+{C{l}_2}_{\left(g\right)}$

at $t=0$ $1$ $0$ $0$

at eqm $1-\alpha$ $\alpha$ $\alpha$

Total moles at equilibrium= $1-\alpha +\alpha +\alpha =1+\alpha$

$P_{SbC{l}_5}=\frac{1-\alpha }{1+\alpha }\times P$ and $P_{SbC{l}_3}=\frac{\alpha }{1+\alpha }\times P$ and $P_{C{l}_2}=\frac{\alpha }{1+\alpha }\times P$

$\Rightarrow K_P=\frac{P_{SbC{l}_3}\times P_{C{l}_2}}{P_{SbC{l}_5}}=\frac{{(\frac{\alpha }{1+\alpha }\times P)}^2}{\frac{1-\alpha }{1+\alpha }\times P}=\frac{{\alpha }^{2}P}{1-{\alpha }^2}$

$\therefore \frac{{\alpha }^{2}P}{1-{\alpha }^2}=1.07$

$\Rightarrow \frac{{\alpha }^2\times 1}{1-{\alpha }^2}=1.07$

$\Rightarrow \alpha =0.718$

$\therefore$ Degree of dissociation is $0.718$ .