The equilibrium constant for the reaction H2 + I2 --> 2HI, is 54 at 425 degrees C. If the equilibrium mixture contains 0.030 M HI and 0.015 M I2, calculate the equilibrium concentration of H2.
Wouldn't Kc be equal to products/Reactants??
To solve this problem, we need to use the equation for the equilibrium constant, which is expressed as:
K = [HI]² / ([H2] x [I2])
We are given the equilibrium constant (K) as 54, and the concentrations [HI] and [I2] as 0.030 M and 0.015 M, respectively. We can substitute these values into the equation and solve for [H2].
Let's assume the equilibrium concentration of H2 is x M.
Using the given values, we have:
54 = (0.030)² / (x * 0.015)
Now, we can solve for x:
54(x * 0.015) = (0.030)²
0.81x = 0.0009
x = 0.0009 / 0.81
x ≈ 0.0011 M
Therefore, the equilibrium concentration of H2 is approximately 0.0011 M.
To calculate the equilibrium concentration of H2, we can use the equation for the equilibrium constant and the given concentrations of HI and I2. Here are the steps to solve the problem:
1. Write the balanced chemical equation for the reaction: H2 + I2 --> 2HI
2. Write the expression for the equilibrium constant (Kc) in terms of the concentrations of the reactants and products:
Kc = [HI]^2 / ([H2] x [I2])
3. Substitute the given values into the equation:
Kc = 54
[HI] = 0.030 M
[I2] = 0.015 M
4. Rearrange the equation to solve for [H2]:
[H2] = [HI]^2 / (Kc x [I2])
5. Plug in the values and calculate:
[H2] = (0.030 M)^2 / (54 x 0.015 M)
[H2] = 0.0069 M
Therefore, the equilibrium concentration of H2 is 0.0069 M.
........H2 + I2 ==> 2HI
E.......x..0.015....0.030
Plug into the Keq expression and solve for x.