Question

If $K_p$ for the reaction $N_2{O}_4\rightleftharpoons 2N{O}_2$ is $0.66$ then what is the equilibrium pressure of $N_2{O}_3$ (Total pressure at equilibrium is $0.5atm)$:

• A. $0.168$
• B. $0.322$
• C. $0.1$
• D. $0.5$

Answer 1

The correct option is A $0.168$

$N_2{O}_4$$\leftrightharpoons 2N{O}_2$

$K_P$=0.66

$P_{eq}$=0.5 atm

$N_2{O}_4$$⟺2N{O}_2$

At t=0 1 0

At t=eq $1-x$ $2x$

Mole fraction of $N_2{O}_4$=$\frac{1-x}{1+x}$

Mole fraction of $N{O}_2$=$\frac{2x}{1+x}$

Partial pressure of $N_2{O}_4$=$\frac{1-x}{1+x}\times 0.5$

Partial pressure of $N{O}_2$=$\frac{2x}{1+x}\times 0.5$

$K_p=\frac{\left[P\right(N{O)}_2{]}^2}{\left[P\right(N{}_2{O}_4)]}0.66=\frac{\frac{2x}{1+x}\frac{2x}{1+x}\times 0.5\times 0.5}{\frac{1-x}{1+x}\times 0.5}0.66=\frac{4x^2\times 0.5}{1-x^2}0.66-0.66x^2=2x^{2}2.66x^2=0.66x=0.5$

Now put the value of x in partial pressure of $N{}_2{O}_4)$

$=\frac{1-0.5}{1+0.5}\times 0.5=\frac{0.5\times 0.5}{1.5}=0.166atm$