Consider the gas-phase reaction, Cl 2 (g) + Br 2 (g) <=> 2 BrCl(g), for which K
ID: 944097 • Letter: C
Question
Consider the gas-phase reaction, Cl2(g) + Br2(g) <=> 2 BrCl(g), for which Kp = 32 at 500 K. If the mixture is analyzed and found to contain 0.45 bar of Cl2, 0.65 bar of Br2 and 3.1 bar of BrCl, describe the situation:
Q < K and more products will be made to reach equilibrium.
Q < K and more reactants will be made to reach equilibrium.
Within 1 decimal place, Q = K and the reaction is at equilibrium
Q > K and more products will be made to reach equilibrium.
Q > K and more reactants will be made to reach equilibrium.
1.Q < K and more products will be made to reach equilibrium.
2.Q < K and more reactants will be made to reach equilibrium.
3.Within 1 decimal place, Q = K and the reaction is at equilibrium
4.Q > K and more products will be made to reach equilibrium.
5.Q > K and more reactants will be made to reach equilibrium.
Explanation / Answer
answer : 5. Q > K and more reactants will be made to reach equilibrium.
Cl2(g) + Br2(g) <=============> 2 BrCl(g)
0.45 0.65 3.1
Q = [BrCl]^2 / [Cl2][Br2]
= 3.1^2 / 0.45 x 0.65
= 32.85
Q = 32.86
Kp = 32
Q > K and more reactants will be made to reach equilibrium.