Question

A mixture of 0.373 atm of NO(g) and 0.31 atm of $C{l}_2\left(g\right)$ is prepared at $500K$. The reaction $2NO\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons 2NOCl\left(g\right)$ takes place. The total pressure at equilibrium is 0.544 atm. Determine $K_p$ for the reaction.

• A. $50.08at{m}^{-1}$
• B. $10.08at{m}^{-1}$
• C. $50.07at{m}^{-1}$
• D. None of the above

Answer 1

The correct option is C $50.07at{m}^{-1}$

The initial pressure are
$P_{NO}=0.373atm$
$PC{l}_2=0.310atm$
$P_{NOCl}=0atm$

The equilibrium pressure are
$P_{NO}=0.373-2xatm$
$P_{C{l}_2}=0.310-xatm$
$P_{NOCl}=2xatm$

Total equilibrium pressure $=(0.373-2x)+(0.310-x)+\left(2x\right)atm.$ But it is equal to 0.544 atm.
$=\left(0.373\right)+(0.310-x)=0.544$
$x=0.139atm$

The equilibrium pressure constant
$K_p=\frac{P_{NOCl}^2}{P_{NO}\times P_{C{l}_2}}$
$K_p=\frac{(2x{)}^2}{(0.373-2x{)}^2\times (0.310-x)}$
$K_p=\frac{\left(2\right(0.139){)}^2}{(0.373-2(0.139){)}^2\times (0.310-0.139)}$

$K_p=50.07at{m}^{-1}$