what is the equillibrium constant of each of the substance in the equillibrium when concentration of ICl was 0.75M 2ICl(g)=I2+Cl2
kc=o.12
You haven't stated the problem very clearly. I assume you mean the equilibrium "concentrations" and not equilibrium constants. You have the Kc given in the problem. Also, I assume that 0.75M is (ICl) initially and not at equilibrium.
..........2ICl ==>I2 + Cl2
I.........0.75.....0.....0
C..........-2x.....x......x
E........0.75-x....x.....x
Substitute the E line nto the Kc expression and solve for x.
To determine the equilibrium constant of each substance in the given equilibrium reaction, we can use the equilibrium law expression:
Kc = [I2][Cl2] / [ICl]^2
Given that the concentration of ICl is 0.75 M and the equilibrium constant (Kc) is 0.12, we need to find the concentrations of I2 and Cl2 to satisfy the equation.
To calculate the concentration of I2 and Cl2, we need to make some assumptions. Let's assume that the change in concentration for ICl is "x" M, and as a result, the change in concentration for I2 and Cl2 is also "x" M.
Since the initial concentration of ICl is 0.75 M and the change in concentration is "x" M, the equilibrium concentration of ICl can be calculated as (0.75 - x) M.
The equilibrium concentration of I2 and Cl2 can be calculated as "x" M.
Now, substituting the equilibrium concentrations into the equilibrium law expression:
Kc = [I2][Cl2] / [ICl]^2
0.12 = (x)(x) / [(0.75 - x)]^2
Simplifying this equation leads to a quadratic equation:
0.12 = x^2 / (0.75 - x)^2
0.12 * (0.75 - x)^2 = x^2
0.09 - 2x(0.75) + x^2 = x^2
0.09 - 1.5x = 0
1.5x = 0.09
x = 0.06
Therefore, the change in concentration of ICl is 0.06 M, and the equilibrium concentration of ICl is (0.75 - 0.06) M = 0.69 M.
The equilibrium concentrations of I2 and Cl2 are both 0.06 M.
Hence, the equilibrium constant of each substance in the equilibrium, when the concentration of ICl is 0.75 M, is:
[K2] = [Cl2] = 0.06 M
[K1] = 0.69 M
Please note that these values are approximations based on the assumptions made.