Question

nitrogen reacts with oxygen to form NO(g) with a K= 0.100. What are the equilibrium, concentration for each chemical if the initial concentration for each chemical if the initial concentrations of N2 (g) is 0.100 M, 02 (g) is 0.500M and NO(g) is 0.100M. Please show quadratic equation.

Answer 1

Explanation:

                            $N_2+O_2\rightleftharpoons 2NO$

Initially Concentration(M)

                       0.100 M  0.500 M      0.100 M

At equilibrium(0.100-x)  (0.500 -x)     (0.100 +2x)

Equilibrium constant of the reaction = $K_c=0.100$

An equilibrium expression for the given reaction is given as:

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$0.100 =\frac{(0.100 +2x)^2}{(0.100 -x)\times (0.500 -x)}$

$0.6 x^2+4.6 x-0.05=0$

On solving this quadratic equation we get:

x = 0.01085

Equilibrium concentration of nitrogen gas"

$[N_2]= (0.100-x)=0.100 M - 0.01085 M= 0.08915 M$

Equilibrium concentration of oxygen gas"

$[O_2]= (0.500-x)=0.500 M - 0.01085 M= 0.48915 M$

Equilibrium concentration of NO gas"

$[N_2]= (0.100+2x)=0.100 M - 2\times 0.01085 M= 0.1217 M$