The equilibrium constant Kc^i is 3.2x10^-34 at 25 degrees C for 2HCl(g) -> H2(g) + Cl2 (g). The equilibrium constant Kc^ii is .10 at 25 degrees C for 2ICl (g) -> Cl2 (g) + I2 (g).
a) Calculate the equilibrium constant Kc for the reaction below at 25 C
Cl2(g) + I2 (g) -> 2ICl (g)
mama
This must be a multipart question sincer the Kc data for HCl is not relevant.
You have 2ICl ==> I2 + Cl2 and you want the reverse of that. Just take the reciprocal of Kc for the forward rxn or
Kc reverse = 1/Kc forward
To find the equilibrium constant, Kc, for the reaction: Cl2(g) + I2(g) -> 2ICl(g), we can use the equation:
Kc = (Kci^ii) / (Kci^i)^2
Given that Kci^i = 3.2x10^-34 and Kci^ii = 0.10, we can substitute these values into the equation:
Kc = (0.10) / (3.2x10^-34)^2
To calculate Kc, we need to square the denominator term:
Kc = (0.10) / (3.2x10^-34)^2
= (0.10) / (3.2x10^-34 x 3.2x10^-34)
= (0.10) / (1.024x10^-67)
To simplify further, we divide 0.10 by 1.024x10^-67:
Kc = 0.10 / 1.024x10^-67
= 9.77x10^65
Therefore, the equilibrium constant Kc for the reaction Cl2(g) + I2(g) -> 2ICl(g) at 25°C is approximately 9.77x10^65.
To calculate the equilibrium constant Kc for the reaction Cl2(g) + I2(g) -> 2ICl(g), we can use the relationship between the equilibrium constants of the given reactions.
The given equilibrium constants are:
Kc^i = 3.2x10^-34 for 2HCl(g) -> H2(g) + Cl2(g)
Kc^ii = 0.10 for 2ICl(g) -> Cl2(g) + I2(g)
The desired reaction is the reverse of reaction (ii), so we will need to take the reciprocal of Kc^ii to calculate the equilibrium constant for the reverse reaction.
Kc^ii_reverse = 1/Kc^ii = 1/0.10 = 10
Now, to calculate the equilibrium constant Kc for the reaction Cl2(g) + I2(g) -> 2ICl(g), we can use the relationship between the equilibrium constants of the forward and reverse reactions.
Kc^forward x Kc^reverse = 1
Kc^forward x 10 = 1
Kc^forward = 1/10
Therefore, the equilibrium constant Kc for the reaction Cl2(g) + I2(g) -> 2ICl(g) is 1/10.