Assume that Kc = 0.0360 at 520°C for the first reaction above. What is the Kc for the second reaction at the same temperature?
2 HI(g)-> H2(g) + I2(g)
H2(g) + I2(g) -> 2 HI(g)
I thought the answer woudl just be negative because all you have to do is switch the products and reactants but its wrong
then like the same for thsi one
Assume that Kc = 0.0130 at 520°C for the first reaction above. What is the Kc for the second reaction at the same temperature?
2 HI(g)->H2(g) + I2(g)
HI(g)->½ H2(g) + ½ I2(g)
why cant you just divide the k by 2?
nevermind i got it
Just to make sure.
K for the reverse direction is 1/K.
K for the 1/2 reaction is sqrt K.
To determine the equilibrium constant (Kc) for the second reaction at the same temperature, we need to understand the relationship between the equilibrium constants of the two reactions.
The equilibrium constant of a chemical reaction is determined by the stoichiometry of the balanced equation. For a balanced equation of the form:
aA + bB ⇌ cC + dD
The equilibrium constant expression, Kc, is given by:
Kc = [C]c[D]d / [A]a[B]b
Given the first reaction:
2 HI(g) ⇌ H2(g) + I2(g)
The equilibrium constant expression for this reaction is:
Kc1 = [H2][I2] / [HI]²
Now, for the second reaction:
H2(g) + I2(g) ⇌ 2 HI(g)
We can see that this is the reverse reaction of the first reaction. The equilibrium constant expression for the second reaction can be obtained by taking the reciprocal of Kc1 because Kc2 is the inverse reaction of Kc1. Therefore:
Kc2 = 1 / Kc1
In this case, if Kc1 = 0.0360, then:
Kc2 = 1 / 0.0360
Kc2 ≈ 27.8
So, the Kc for the second reaction at the same temperature is approximately 27.8.