Consider the equilibrium between SbCl5, SbCl3 and Cl2. SbCl5(g)<--------> SbCl3(
ID: 941524 • Letter: C
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
Explanation / Answer
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: