Question

Consider the reaction:

N2(g) + O2(g) ⇌ 2NO(g)

EQUIL. 0.200 atm 0.100 atm 1.35 x 10^-4 atm

Calculate the equilibrium pressure of NO(g) AFTER the addition of 0.400 atm of O2(g).

A. 7.82 x 10-4 atm

B. 8.43 x 10-4 atm

C. 3.02 x 10-4 atm

D. 5.02 x 10-4 atm

E. 2.23 x 10-4 atm

Answer 1

The equilibrium constant $K_P= \dfrac {P^2_{NO}}{P_{N2}P_{O2}}$

$K_P= \dfrac {\left ( 1.35 \times 10^{-4} \right )^2}{0.200 \times 0.100}$

$K_P=9.11 \times 10^{-7}$

	$N_2$	$O_2$	$NO$
Initial pressure (atm)	0.200	0.100	$1.35 \times 10^{-4}$
Change in pressure (atm)	-x	0.400-x	2x
Equilibrium pressure (atm)	0.200-x	0.500-x	$1.35 \times 10^{-4}+2x$

The equilibrium constant $K_P= \dfrac {P^2_{NO}}{P_{N2}P_{O2}}$

$9.11 \times 10^{-7}= \dfrac {\left ( 1.35 \times 10^{-4}+2x \right )^2}{\left ( 0.200-x \right ) \times \left ( 0.100-x \right )}$

$9.11 \times 10^{-7}= \dfrac {\left ( 1.35 \times 10^{-4}+2x \right )^2}{\left ( 0.200-x \right ) \times \left ( 0.500-x \right )}$

Since the value of Kp is very small, we equate 0.200-x to 0.200 and 0.500-x to 0.500

$9.11 \times 10^{-7}= \dfrac {\left ( 1.35 \times 10^{-4}+2x \right )^2} {0.200 \times 0.500}$

$9.11 \times 10^{-8} =\left ( 1.35 \times 10^{-4}+2x \right )^2$

$1.35 \times 10^{-4}+2x=0.000301869$

$x = 8.34 \times 10^{-5}$

Hence, the equilibrium pressure of NO(g) AFTER the addition of 0.400 atm of O2(g).

$=1.35 \times 10^{-4}+2x = 1.35 \times 10^{-4}+\left ( 2 \times 8.34 \times 10^{-5} \right )=3.02 \times 10^{-4} \: atm$

Thus, the option (C) is the answer.