Calculate the equilibrium concentration of NO(g).

Question

Consider the following equilibrium with K_c = 1.6 × 10^–5:

  2NOCl(g)  2NO(g) + Cl₂(g)

 2.00 mole of pure NOCl and 1.808 mole of pure Cl₂ are placed in a 2.00-L container. 

{Remember: you will need to first calculate the molar concentrations. Also notice that the "x is small" approximation explained in section 12.6 will simplify this calculation).

a. 2.10 * 10^-3 M

b. 9.04 * 10^-1 M

c. 1.11 M

d. 5.95 * 10^-3 M

e. 4.21 * 10^-3 M

Answer 1

[NOCl]initial=2mol/2L=1M

[Cl2]initial=1.808mol/2L=0.904M

[NO]initial=0M

2NOCl(g)↔2NO(g)+Cl2(g)

1..................0.............0.904

-2x..............+2x...........+x

1-2x............2x..........0.904+x

Remember: you will need to first calculate the molar concentrations. Also notice that the "x is small" approximation explained in section 12.6 will simplify this calculation

"-2x" and "+x" are very small and ignored when solving for x, so that only the "2x" is present to solve for x.

Basically, you don't do the quadratic equation and ignored the "+x" or "-x" when solving for x, but then put it back to solve for the equilibrium concentrations.

1.6E-5=( [2x]^2 [0.904])/([1-2x]^2)

1.6E-5=( [2x]^2 [0.904])/([1]^2)

x=2.10E-3

What're you finding the equilibrium concentration of? you didn't specify.

[NOCl]eq=1-2x=1-2(0.00210)=0.9958

[NO]eq=2x=2(0.00210)=0.00420

[Cl2]eq=0.904+x=0.904+0.00210=0.9061