Question

In the reaction $COC{l}_2\left(g\right)\rightleftharpoons CO\left(g\right)+C{l}_2\left(g\right)$ at ${550}^{o}C$, when the initial pressure of CO & $C{l}_2$ are 250 and 280 mm of Hg respectively. The equilibrium pressure is found to be 380 mm of Hg. Calculate the degree of dissociation of $COC{l}_2$ at 1 atm. What will be the extent of dissociation, when $N_2$ at a pressure of 0.4 atm is present and the total pressure in 1 atm?

• A. 0.32 and no change
• B. 0.32 and 0.40
• C. 0.32 and 0.42
• D. In presence of $N_2$ dissociation cannot take place

Answer 1

The correct option is B 0.32 and 0.40

$COC{l}_2\left(g\right)\rightleftharpoons CO\left(g\right)+C{l}_2\left(g\right)$
- 250 280 Initial pressure in mm of Hg
x 250-x 280-x Equilibrium pressure in mm of Hg
$x+250-x+280-x=380$
$x=150$ mm of Hg

$K_p=\frac{P_{CO}\cdot P_{C{l}_2}}{P_{COC{l}_2}}$
$K_p=0.114$
$K_p=\frac{{\alpha }^2\cdot p}{1-{\alpha }^2}$
$0.114=\frac{{\alpha }^2\cdot 1}{1-{\alpha }^2}$
Hence $\alpha =0.32$
In presence of $N_2$ (constant pressure process)
$K_p=\frac{{\alpha }^2\times 0.6}{1-{\alpha }^2}$
$0.114=\frac{{\alpha }^2\times 0.6}{1-{\alpha }^2}$
$\alpha =\sqrt{\frac{0.114}{0.714}}$
$\alpha =0.4$
$\alpha$ increases from 0.32 to 0.4