At a particular temperature, K = 1.00 102 for the following reaction.
H2(g) + I2(g) 2 HI(g)
In an experiment, 1.23 mol H2, 1.23 mol I2, and 1.23 mol HI are introduced into a 1.00 L container. Calculate the concentrations of all species when equilibrium is reached.
To calculate the equilibrium concentrations of all species, we can use the expression for the equilibrium constant (K) and the stoichiometry of the reaction.
First, let's define the initial concentrations of H2, I2, and HI as [H2]0, [I2]0, and [HI]0, respectively. In this case, [H2]0 = 1.23 mol/L, [I2]0 = 1.23 mol/L, and [HI]0 = 1.23 mol/L.
Let's assume that at equilibrium, the concentrations of H2, I2, and HI are [H2]eq, [I2]eq, and [HI]eq, respectively.
According to the balanced equation, the stoichiometric coefficients for H2, I2, and HI are 1, 1, and 2 respectively. So, the molar changes for H2, I2, and HI can be expressed as follows:
[H2]eq = [H2]0 - Δn(H2)
[I2]eq = [I2]0 - Δn(I2)
[HI]eq = [HI]0 + Δn(HI)
where Δn(H2) = -1, Δn(I2) = -1, and Δn(HI) = +2.
Now, let's calculate the value of Δn:
Δn = (2 - 2) - (1 - 1) = 0
Since Δn = 0, the concentrations of all species remain unchanged at equilibrium.
Therefore, the equilibrium concentrations of H2, I2, and HI are:
[H2]eq = [H2]0 = 1.23 mol/L
[I2]eq = [I2]0 = 1.23 mol/L
[HI]eq = [HI]0 = 1.23 mol/L
So, the concentrations of all species remain the same as their initial values when equilibrium is reached.
To calculate the concentrations of all species when equilibrium is reached, we need to use the equation for the equilibrium constant (K) and the given initial amounts of reactants.
1. First, write out the balanced chemical equation:
H2(g) + I2(g) ⇌ 2 HI(g)
2. Next, use the given value of K (K = 1.00 x 10^2) to set up the equilibrium expression:
K = [HI]^2 / ([H2] * [I2])
3. Now, substitute the given initial amounts into the equilibrium expression. Since the initial volume of the container is 1.00 L, the initial concentrations are equal to the initial amounts of the substances divided by the volume (1.23 mol / 1.00 L):
[H2] = 1.23 mol / 1.00 L = 1.23 M
[I2] = 1.23 mol / 1.00 L = 1.23 M
[HI] = 1.23 mol / 1.00 L = 1.23 M
4. Plug these values into the equilibrium expression and solve for [HI] using algebra:
1.00 x 10^2 = ([HI]^2) / (1.23 * 1.23)
1.00 x 10^2 = ([HI]^2) / 1.5129
Multiply both sides by 1.5129:
1.00 x 10^2 * 1.5129 = [HI]^2
151.29 = [HI]^2
Now take the square root of both sides to solve for [HI]:
[HI] = √(151.29) ≈ 12.3 M
5. Substitute the value of [HI] back into the equilibrium expression to solve for [H2] and [I2]:
1.00 x 10^2 = (12.3^2) / ([H2] * [I2])
Rearrange the equation:
[H2] * [I2] = (12.3^2) / (1.00 x 10^2)
Multiply both sides by ([H2] * [I2]):
[H2] * [I2] * (1.00 x 10^2) = (12.3^2)
Now, substitute the values of [H2] = 1.23 M and [I2] = 1.23 M:
(1.23) * (1.23) * (1.00 x 10^2) = (12.3^2)
Simplify:
151.29 = 151.29
Therefore, the concentrations of all species at equilibrium are:
[H2] = 1.23 M
[I2] = 1.23 M
[HI] = 12.3 M
x =.9225
and when you substitute (1.23-.9225) makes H2 and I2=.3075
(1.23+2*.9225)=3.075
(H2) = 1.23/1L = 1.23M
(I2) = 1.23/1L = 1.23M
(HI) = 1.23/1 = 1.23M
First you want to calculate Q = (HI)^2/(H2)(I2) = (1.23)^2(1.23)(1.23) = 1.00 which means HI is too small and H2 and I2 are too large so the reaction will go to the right.
............H2 + I2 ===> 2HI
initial....1.23..1.23.....1.23
change......-x....-x.....2x
equil.....1.23-x..1.23-x..1.23+2x
Substitute from the ICE chart into K expression and solve for x.