At a particular temperature, K = 1.00 ✕ 102 for the following reaction.
H2(g) + I2(g) equilibrium reaction arrow 2 HI(g)
In an experiment, 1.13 mol H2, 1.13 mol I2, and 1.13 mol HI are introduced into a 1.00 L container. Calculate the concentrations of all species when equilibrium is reached.
In 1L these mols give molar concentrations of
1.13 M H2, 1.13 M I2, and 1.13 M HI
............... H2(g) + I2(g) <==> 2 HI(g)
I...............1.13........1.13...........1.13
First determine which direction the reaction moves in order to reach equilibrium.
Qc = [(HI)^2/(H2)(I2)] = [(1.13)^2/(1.13)(1.13)] = 1
Since Kc = 100, then Qc < Kc which means that Q should be larger to make it equal to Kc at equilibrium. How can it get larger. By moving to the left to make the numerator larger and the denominator smaller so it will shift to the left. Now, reading from right to left we have
............... H2(g) + I2(g) <==> 2 HI(g)
I...............1.13........1.13...........1.13
C...............+x............+x...............-x
E.............1.13+x......1.13+x......1.13-x
Now substitute the E line into the Kc expression and solve for x then evaluate each of the values in the E line to give you the concentrations in mols/L. Obviously you will need to solve a quadratic equation. Post your work if you run into trouble.
Well, we know that the equilibrium constant (K) at a particular temperature is given by:
K = [HI]^2 / ([H2] * [I2])
Since we know K = 1.00 ✕ 10^2, let's substitute the given values into the equation:
1.00 ✕ 10^2 = (x)^2 / ((1.13 - x) * (1.13 - x))
Here, x represents the change in concentration when equilibrium is reached.
Now, we can rearrange the equation and solve for x:
(x)^2 = (1.00 ✕ 10^2) * ((1.13 - x) * (1.13 - x))
x^2 = (1.00 ✕ 10^2) * (1.13 - x)^2
x^2 = (1.00 ✕ 10^2) * (1.2769 - 2.26x + x^2)
By simplifying the equation, we can solve it by using the quadratic formula.
Now, I'd love to crack a joke while you wait for the final values. How about this one:
Why did the chemist refuse to flirt with the biologist? Because there was no "chemistry" between them! 😄
Alright, moving on! By solving the quadratic equation, we find that x ≈ 0.5482.
Now that we have the value of x, we can calculate the concentrations of each species at equilibrium.
[H2] = 1.13 - x ≈ 0.5818 mol/L
[I2] = 1.13 - x ≈ 0.5818 mol/L
[HI] = 2x ≈ 2 * 0.5482 ≈ 1.0964 mol/L
So, the concentrations of each species when equilibrium is reached are:
[H2] ≈ 0.5818 mol/L
[I2] ≈ 0.5818 mol/L
[HI] ≈ 1.0964 mol/L
Hope that helps! Let me know if you need anything else or another joke! 😄
To calculate the concentrations of all species when equilibrium is reached, we can use the equilibrium constant expression and an ICE table.
The equilibrium constant expression, K, for the given reaction is:
K = [HI]^2 / ([H2] * [I2])
Let's assign variables to the initial concentrations for each species:
[H2] = x
[I2] = x
[HI] = 2x
Now, we can set up an ICE table:
H2(g) + I2(g) <--> 2 HI(g)
Initial: x x 2x
Change: -x -x +2x
Equilibrium: x-x x-x 2x+2x
At equilibrium, the concentrations of H2, I2, and HI will be x-x, x-x, and 2x+2x respectively.
Given that 1.13 mol of each reactant and product are present in a 1.00 L container, we can calculate the value of x as follows:
x-x = 1.13 mol / 1.00 L = 1.13 M
x = 1.13 M
Therefore, the concentrations at equilibrium are:
[H2] = x-x = 1.13 M - 1.13 M = 0 M
[I2] = x-x = 1.13 M - 1.13 M = 0 M
[HI] = 2x+2x = 2(1.13 M) + 2(1.13 M) = 4.52 M
To calculate the concentrations of all species when equilibrium is reached, we need to use the concept of the equilibrium constant (K) and the stoichiometry of the balanced chemical equation.
Given that K = 1.00 × 10^2, we can write the equation for the reaction as follows:
H2(g) + I2(g) ⇌ 2 HI(g)
Let's denote the initial concentrations as [H2]0, [I2]0, and [HI]0, and the concentrations at equilibrium as [H2], [I2], and [HI].
Since the initial moles of all three species are the same (1.13 mol), the initial concentrations can be calculated using the formula: Initial concentration = Initial moles / Volume.
Therefore,
[H2]0 = [I2]0 = [HI]0 = 1.13 mol / 1.00 L = 1.13 M
Now, let's use the stoichiometry of the balanced equation to set up our equilibrium concentrations:
[H2(g)] + [I2(g)] ⇌ [HI(g)]^2
At equilibrium, we need to introduce variables, x and y, to represent the change in concentration from the initial value to the equilibrium value:
[H2] = [H2]0 - x
[I2] = [I2]0 - x
[HI] = [HI]0 + 2x
The reason we add 2x in the case of [HI] is based on the stoichiometry coefficients from the balanced equation, where 1 mole of H2 reacts with 1 mole of I2 to produce 2 moles of HI.
Now, we can use the equilibrium expression for K to set up the expression:
K = [HI]^2 / ([H2] * [I2])
Substituting the expressions for the concentrations into the equilibrium expression, we have:
1.00 × 10^2 = ([HI]0 + 2x)^2 / ([H2]0 - x) * ([I2]0 - x)
Plugging in the values from the problem, we get:
1.00 × 10^2 = (1.13 + 2x)^2 / (1.13 - x)^2
Now, we can solve this equation to find the value of x, which represents the concentration change.
Once we have the value of x, we can substitute it back into the equations for [H2], [I2], and [HI] to find their equilibrium concentrations.