Consider the gas-phase reaction, Cl2(g) + Br2(g) 2 BrCl(g), for which Kp = 32 at
ID: 895437 • 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 0.7 bar of Cl2, 0.7 bar of Br2 and 4.0 bar of BrCl, describe the situation:
1. Q < K and more reactants will be made to reach equilibrium.
2. Q > K and more reactants will be made to reach equilibrium.
3. Q < K and more products will be made to reach equilibrium.
4. Q > K and more products will be made to reach equilibrium.
5. Within 1 decimal place, Q = K and the reaction is at equilibrium
Explanation / Answer
I will suppose reaction goes:
Cl2 + Br2 ---> 2BrCl
Kp = 32 at 500K
PCl = 0.7
PBr = 0.7
PBrCl = 4
Kp = PBrCl^2 / (PCl2)(PBr2)
The actual value of Q is:
Q = PBrCl^2 / (PCl2)(PBr2) = 4^2 / (0.7)(0.7) = 32.653
Technically speaking
Q > K, therefore expect more products than reactants in the equilibrium
BUT, there is a very special note in 5.)
Within 1 decimal place, we can ssume Q=K, which is the case since 32.0 is almost 32.6
Therefore, this is true, you can assume this is in equilibrium (choose 5)