the initial concentration of each component was: 0.100 M H2(g), 0.100M I2(g), and 0.050 M HI(g) . The Keq = 50.2. Calculate the concentration of each component when equilibrium has been reached. The balanced chemical equation for this reactions is: H2(g) + I2(g) 2HI(g)
To solve this problem, we will use the balanced chemical equation and the given initial concentrations to calculate the equilibrium concentrations of each component.
The balanced chemical equation is: H2(g) + I2(g) ⇌ 2HI(g)
Let's assume that at equilibrium, the concentration of H2(g) is x M, the concentration of I2(g) is x M, and the concentration of HI(g) is 2x M.
Using the given initial concentrations:
[H2] = 0.100 M
[I2] = 0.100 M
[HI] = 0.050 M
At equilibrium, the concentrations can be represented as:
[H2] = 0.100 - x M
[I2] = 0.100 - x M
[HI] = 0.050 + 2x M
According to the equation Keq = [HI]^2 / ([H2] * [I2]), we can write:
50.2 = (0.050 + 2x)^2 / ((0.100 - x) * (0.100 - x))
Now, we'll solve this equation to find the value of x.
50.2 * (0.100 - x) * (0.100 - x) = (0.050 + 2x)^2
Simplifying:
5.02 * (0.100 - x)^2 = (0.050 + 2x)^2
Taking the square root of both sides:
√(5.02 * (0.100 - x)^2) = 0.050 + 2x
Simplifying further:
√(5.02 * (0.010 - 0.200x + x^2)) = 0.050 + 2x
5.02 * (0.010 - 0.200x + x^2) = (0.050 + 2x)^2
Expanding and rearranging:
0.0501 - 0.10x + 5.02x^2 = 0.0025 + 0.205x + 4x^2
Combining like terms:
0.0005 + 0.30x = 1.02x^2
Now, we'll rearrange the equation to solve for x:
1.02x^2 - 0.30x - 0.0005 = 0
Using the quadratic formula:
x = [-(-0.30) ± √((-0.30)^2 - 4(1.02)(-0.0005))] / (2 * 1.02)
Simplifying:
x = (0.30 ± √(0.09 + 0.00408)) / 2.04
Calculating:
x = (0.30 ± √0.09408) / 2.04
Now, we have two possible values for x. Let's calculate them:
1. x = (0.30 + √0.09408) / 2.04 ≈ 0.100 M
2. x = (0.30 - √0.09408) / 2.04 ≈ -0.003 M (Ignore this value because it is not physically meaningful)
Since x represents the concentration of H2, the concentration of H2 at equilibrium is approximately 0.100 M.
Now we can calculate the equilibrium concentrations of the other components:
[H2] = 0.100 - x = 0.100 - 0.100 = 0 M
[I2] = 0.100 - x = 0.100 - 0.100 = 0 M
[HI] = 0.050 + 2x = 0.050 + 2(0.100) = 0.250 M
Therefore, at equilibrium:
[H2] ≈ 0 M
[I2] ≈ 0 M
[HI] ≈ 0.250 M
To calculate the concentration of each component at equilibrium, we'll use the concept of equilibrium constant (Keq) and the stoichiometry of the reaction.
The given balanced chemical equation is:
H2(g) + I2(g) ⇌ 2HI(g)
According to the stoichiometry of the balanced equation, 1 mole of H2 reacts with 1 mole of I2 to produce 2 moles of HI.
Let's assume 'x' as the change in the concentration (in moles) of H2 and I2, and '2x' as the change in the concentration of HI. At equilibrium, the initial concentration minus the change will give us the concentration of each component.
Using the given initial concentrations, let's denote the change in concentration for H2 and I2 as '-x' and for HI as '-2x'.
The equilibrium concentration of H2 will be:
[H2]eq = [H2]initial - x = 0.100 - x
The equilibrium concentration of I2 will be:
[I2]eq = [I2]initial - x = 0.100 - x
The equilibrium concentration of HI will be:
[HI]eq = [HI]initial + 2x = 0.050 + 2x
Now, we can set up the expression for the equilibrium constant, Keq:
Keq = [HI]eq² / ([H2]eq * [I2]eq)
Substituting the equilibrium concentrations, we have:
Keq = (0.050 + 2x)² / ((0.100 - x) * (0.100 - x))
Since Keq = 50.2, we can solve the equation to find the value of 'x'.
50.2 = (0.050 + 2x)² / ((0.100 - x) * (0.100 - x))
Simplifying and rearranging the equation, we get a quadratic equation:
250.5 - 10.02x + 4x² = 0
Solving this quadratic equation will give us the value of 'x', which represents the change in concentration at equilibrium.
Once we find the value of 'x', we can substitute it into the equilibrium expressions to calculate the equilibrium concentrations of each compound:
[H2]eq = 0.100 - x
[I2]eq = 0.100 - x
[HI]eq = 0.050 + 2x
These values will give us the concentration of each component when equilibrium is reached.
First, calculate Q
(HI)^2/(H2)(I2) = (0.05)^2/(O>1)^2 = 0.25 which is too small; therefore the reaction will move to the right.
..........H2 + I2 ==> 2HI
initial..0.1...0.1....0.05
change..-x....-x.....+x
equil....0.1-x..0.1-x...0.05+x
Set up Kc = (HI)^2/(H2)(I2)
Solve for x, 0.1-x and 0.05+x