The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant.
2HBr(g) <--> H2(g) + Br2(g) Kc=0.322
H2(g) + Br2(g) <--> 2HBr(g) Kc= ?
Kc for rxn 2 is 1/Kc for rxn 1.
To determine the value of the missing equilibrium constant, we can use the concept of equilibrium constant reciprocals.
Given the equilibrium constant for the first reaction:
2HBr(g) <--> H2(g) + Br2(g) Kc = 0.322
We want to find the equilibrium constant for the second reaction:
H2(g) + Br2(g) <--> 2HBr(g) Kc = ?
The second reaction is the reverse reaction of the first reaction. Therefore, the equilibrium constant for the second reaction is the reciprocal of the equilibrium constant for the first reaction.
1/(0.322) = 3.105
So, the equilibrium constant for the second reaction is 3.105.
To determine the value of the missing equilibrium constant Kc, we can use the relationship between the forward and reverse reactions. The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.
For the reaction:
2HBr(g) <--> H2(g) + Br2(g) Kc=0.322
The forward reaction can be written as:
H2(g) + Br2(g) <--> 2HBr(g)
Since we are looking for the equilibrium constant for this reverse reaction, we can use the reciprocal of the given equilibrium constant:
Kc(reverse) = 1 / Kc(forward)
Therefore:
Kc(reverse) = 1 / 0.322
Calculating this value gives:
Kc(reverse) = 3.11
So, the missing equilibrium constant Kc is approximately 3.11.