The air pollutant NO is produced in automobile engines from the high-temperature

Question

The air pollutant NO is produced in automobile engines from the high-temperature reaction:
N2(g)+O2(g)2NO(g) Kc=1.7×103at 2300 K.

Part A

Calculate the equilibrium concentration of [N2] if the initial concentrations are 2.25 M N2 and 0.53 M O2. (This N2/O2 concentration ratio is the ratio found in air.)

Express your answer to three significant figures and include the appropriate units.

The air pollutant NO is produced in automobile engines from the high-temperature reaction:
N2(g)+O2(g)2NO(g) Kc=1.7×103at 2300 K.

Part A

Calculate the equilibrium concentration of [N2] if the initial concentrations are 2.25 M N2 and 0.53 M O2. (This N2/O2 concentration ratio is the ratio found in air.)

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

ICE Table:

[N2] [O2] [NO]

initial 2.25 0.53 0

change -1x -1x +2x

equilibrium 2.25-1x 0.53-1x +2x

Equilibrium constant expression is

Kc = [NO]^2/[N2]*[O2]

0.0017 = (4*x^2)/((2.25-1*x)(0.53-1*x))

0.0017 = (4*x^2)/(1.193-2.78*x + 1*x^2)

2.027*10^-3-4.726*10^-3*x + 1.7*10^-3*x^2 = 4*x^2

2.027*10^-3-4.726*10^-3*x-3.998*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = -3.998

b = -4.726*10^-3

c = 2.027*10^-3

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.244*10^-2

roots are :

x = -2.312*10^-2 and x = 2.193*10^-2

since x can't be negative, the possible value of x is

x = 2.193*10^-2

At equilibrium:

[N2] = 2.25-1x = 2.25-1*0.02193 = 2.22807 M

Answer: 2.23 M

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