Question

Consider the equilibrium between SbCl5, SbCl3 and Cl2.

SbCl5(g)<--------> SbCl3(g) + Cl2(g) K = 2.36×10-2 at 552 K

The reaction is allowed to reach equilibrium in a 6.80-L flask. At equilibrium, [SbCl5] = 0.359 M, [SbCl3] = 9.20×10-2 M and [Cl2] = 9.20×10-2 M.

(a) The equilibrium mixture is transferred to a 13.6-L flask. In which direction will the reaction proceed to reach equilibrium?

(b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 13.6-L flask.

[SbCl5] =______M

[SbCl3] =______M

[Cl2] =________M

Answer 1

SbCl5(g)<--------> SbCl3(g) + Cl2(g) K = 2.36×10-2 at 552 K

[SbCl5] = 0.359 M,

M1V1=M2V2

0.359 M * 6.80 L = M2 * 13.6

M2 = 0.1795 M

[SbCl3] = 9.20×10-2 M

M1V1=M2V2

9.20×10-2 M * 6.80 L = M2 * 13.6

M2 = 0.046 M

and [Cl2] = 9.20×10-2 M.

M1V1=M2V2

9.20×10-2 M * 6.80 L = M2 * 13.6

M2 = 0.046 M

Now calculate the reaction quotient as follows;

Q = 0.046 M * 0.046 M / 0.1795 M

Q= 0.0118

K = 2.36×10-2 at 552 KQ < K

When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

The relation between equilibrium consatnt K p and reaction quotient Q as follows:

Q = K

When Q=K, the system is at equilibrium and there is no shift to either the left or the right.

Q < K

When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

Q > K

When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants. For Q>K: