Kc for the reaction of hydrogen and iodine to produce hydrogen iodide,
H2(g)+I2(g)f 2HI(g)
is 54.3 at 430C. What will the concentrations be at equilibrium if we start with 0.240M concentrations of both H2 and I2?
To determine the concentrations of the reactants and products at equilibrium, we can use the equilibrium expression and the given value of Kc.
The equilibrium expression for the reaction is:
Kc = [HI]^2 / ([H2] * [I2])
Given:
Initial concentration of H2 = 0.240M
Initial concentration of I2 = 0.240M
Kc = 54.3
Let's assume the change in concentration for HI is 'x' at equilibrium. Since the stoichiometric coefficient for HI is 2, the concentrations of H2 and I2 will decrease by 'x', and the concentration of HI will increase by '2x'.
So, at equilibrium:
[H2] = 0.240 - x
[I2] = 0.240 - x
[HI] = 2x
Substituting these values into the equilibrium expression:
54.3 = (2x)^2 / ((0.240 - x) * (0.240 - x))
Now, we can solve this equation to find the value of 'x', which represents the change in concentration at equilibrium.
Simplifying the equation:
54.3 = 4x^2 / ((0.240 - x)^2)
Multiply both sides by ((0.240 - x)^2):
54.3 * ((0.240 - x)^2) = 4x^2
Expanding and rearranging the equation:
13.032 * (0.0576 - 0.480x + x^2) = 4x^2
Multiplying out the terms:
0.748032 - 6.99744x + 13.032x^2 = 4x^2
Rewriting it as a quadratic equation:
9.032x^2 - 6.99744x + 0.748032 = 0
Now, we can solve this quadratic equation using the quadratic formula to find the value of 'x'.
To determine the concentrations at equilibrium, we can use the equilibrium expression and the initial concentrations of H2 and I2.
The equilibrium expression for the reaction is:
Kc = [HI]^2 / [H2][I2]
Given:
Kc = 54.3
Initial concentration of H2 = 0.240 M
Initial concentration of I2 = 0.240 M
Since the stoichiometric coefficients for H2, I2, and HI are all 1, we can assume that the change in concentrations from the initial to the equilibrium state will be the same.
Let x be the change in concentration of HI at equilibrium.
At equilibrium, the concentration of H2 will be (0.240 - x) M.
The concentration of I2 will also be (0.240 - x) M.
Substituting these values into the equilibrium expression, we get:
54.3 = (x^2) / [(0.240 - x)(0.240 - x)]
Now, we can solve this equation to find the value of x.
54.3 = (x^2) / [(0.240 - x)^2]
Multiplying both sides by [(0.240 - x)^2], we get:
(0.240 - x)^2 = (x^2) / 54.3
Expanding and rearranging the equation, we get:
0.0576 - 0.480x + x^2 = x^2 / 54.3
Rearranging again, we get:
x^2 / 54.3 - x^2 = 0.0576 - 0.480x
Combining like terms, we get:
x^2 (1/54.3 - 1) = 0.0576 - 0.480x
Simplifying further, we get:
x^2 (1 - 54.3) = 0.0576 - 0.480x
x^2 (-53.3) = 0.0576 - 0.480x
x^2 = (0.0576 - 0.480x) / (-53.3)
Solving this equation, we find the value of x.
Set up an ICE chart.
............H2 + I2 ==> 2HI
initial.0.240...0.240....0
change......-x....-x.....2x
equil...0.240-x..0.240-x..2x
Substitute into Kc expression.
Kc = 54.3 = (HI)^2/(H2)(I2)
Solve for x and evaluate 0.240-x