Question

For the dissociation of phosgene gas, the value of $K_p$ at $100$ is $6.7\times {10}^{-9}atm$. The fraction of phosgene dissociated at this temperature when 1 mole of phosgene is placed in 100 litre vessel containing nitrogen at partial pressure of 1 atm is $X\times {10}^{-4}$. $100X$ is__________.

Answer 1

	$COC{l}_2$	$CO$	$C{l}_2$
Initial number of moles	1	0	0
Equilibrium number of moles	$1-x$	$x$	$x$

$K_P=K_C(RT{)}^{\Delta n}$
$6.7\times {10}^{-9}=K_C(0.0821\times 373)$
$K_C=2.19\times {10}^{-10}$
$\frac{\frac{x^2}{10000}}{\frac{1-x}{100}}\times \times {2.1910}^{-10}$
$\frac{x}{100}=1.479\times {10}^{-5}$
$x=1.479\times {10}^{-4}=X\times {10}^{-4}$
$X=1.48$
$100X=148$