Question

The decomposition of NOCl: 2NOCl (g) <---> Cl2 (g) + 2NO (g) has an equilibrium constant of K=1.6*10^-5 at 35 degrees C. if the initial concentration of NOCl is 0.500M, what is the equilibrium concentration of NO?

A. 1.00*10^-4M

B. 2.00*10^-2M

C. 1.00*10^-2M

D. 2.3*10^-3M

When sodium is added to the reaction vessel, it scavenges the chlorine gas in an exothermic reaction. What will the impact of adding sodium be on the equilibrium of the decomposition of NOCl?

A. Shifts towards products

B. Shifts towards the reactants

C. There will be no change because the sodium isnt participating in the chemical reaction

D. This question cannot be answered with the information provided

Answer 1

Q1

initially

[NOCl] = 0.5

[Cl2] = 0

[NO] =0

in equilbirium

[NOCl] = 0.5 - 2x

[Cl2] = 0 + x

[NO] =0 + 2x

substitutein K

1.6*10^-5 = (2x)^2(x) / (0.5-2x)^2

(1.6*10^-5)(0.5^2-x+4x^2) = 4*x^3

4*10^-6- 1.6*10^-5 x + 0.000064x^2 = 4x^3

4x^3 - 0.000064x^2 + 0.000016x - 0.000004 = 0

x = 0.00987

[NOCl] = 0.5 - 2*0.00987 = 0.48026 M

[Cl2] = 0 + 0.00987 = 0.00987 M

[NO] =0 + 2*0.00987 = 0.01974 M

Q2

if we add Na, then Cl2 decreases, which implies that NOCl must form even more Cl2, that is, shift to the right/prducts