Question

15.

This question has multiple parts. Work all the parts to get the most points. At 35°C, K-1.1 x 10° for the reaction 2 NOCI(g) 근 2N0(g) + Cl2 (g) Calculate the concentrations of all species at equilibrium for each of the following original mixtures a3.8 moles of pure NOCl in a 2.0-L flask [NOCI] = NO] [C12] = b1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask NOCI NO Numeric input field 3.8 mole of NOCl and 1.0 mole of Cl2 in a 1.0-L flask NOCI [NO] = Cl Cc

Answer 1

a)

2NOCl(g) <------> 2NO(g) + Cl2(g)

K = [NO]²[Cl2]/[NOCl]² = 1.1×10^-5

Initial concentration

[NOCl] = (3.8mol/2L)×1L = 1.9

[NO] = 0

[Cl] = 0

change in concentration

[NOCl] = -2x

[NO] = +2x

[Cl] = +x

Equillibrium concentration

[NOCl] = 1.9 -2x

[NO] = 2x

[Cl2] = x

So,

( 2x)²x /(1.9 -2x)² = 1.1×10^-5

solving for x

x = 0.02117

Therefore, at equillibrium

[NOCl] = 1.9 - 2(0.02117) = 1.8577M

[NO] = 2×0.02117 = 0.04234M

[Cl] = 0.02117M

b)

Initial concentration

[NOCl] = 1.0mol/1L = 1.000M

[NO] = 1.0mol/1L = 1.000M

[Cl2] = 0

change in concentration

[NOCl] = -2x

[NO] = +2x

[Cl2] = +x

Equillibrium concentration

[NOCl] = 1.000 - 2x

[NO] = 1.000 + 2x

[Cl2] = x

so,

(1.000 + 2x)²(x) / (1.000 - 2x)²= 1.1×10^-5

^{solving for x}

  x = 0.000010999.

Therefore, at equillibrium

[NOCl] = 1.000 -2(0.000010999) = 1.000M

[NO] = 1.000 + 2(0.000010999) = 1.000M

[ Cl2] = 0.000010999M

c)

Initial concentration

[NOCl] = 3.8mol/1L = 3.800M

[NO] =0

[Cl2] = 1.0mol/1L = 1.000M

change in concentration

[NOCl] = -2x

[NO] = +2x

[Cl2] = +x

Equillibrium concentration

[NOCl] = 3.800 - 2x

[NO] = 2x

[Cl2] = 1.000 +x

so,

(2x)²(1.000 +x) /(3.800 - 2x)² = 1.1× 10^-5

solving for x

x = 0.006261

Therefore, at equillibrium

[NOCl] = 3.800 - (2×(0.006261)) = 3.7875M

[NO] = 2(0.006261) = 0.0125M

[Cl2] = 1.000+ 0.006261= 1.0063M