At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.
At this temperature, 0.300 mol of H2 and 0.300 mol of I2 were placed in a 1.00-L container to react. What concentration of HI is present at equilibrium?
I've done two of this type for you; try this on your own and see what you can do. Post your work if you need further assistance.
.653
To answer this question, we can use the information provided along with the concept of equilibrium constant (Kc) and stoichiometry.
Step 1: Write the balanced equation for the reaction:
H2 + I2 ⇌ 2HI
Step 2: Set up an ICE (Initial, Change, Equilibrium) table to keep track of the moles/concentrations of each species involved in the reaction.
| H2 | I2 | HI |
-----------------------------------------------------------
Initial | 0.300 | 0.300 | 0 (since none react)
Change | -x (decreases) | -x (decreases) | +2x (increases)
Equilibrium | 0.300 - x | 0.300 - x | 2x
Step 3: Write the expression for the equilibrium constant (Kc):
Kc = [HI]^2 / ([H2] * [I2])
Step 4: Substitute the equilibrium concentrations into the Kc expression:
Kc = (2x)^2 / ((0.300 - x) * (0.300 - x))
= 4x^2 / (0.09 - 0.6x + x^2)
Step 5: Set up the equation and solve for x:
Kc = 53.3 (given)
53.3 = 4x^2 / (0.09 - 0.6x + x^2)
Since this is a quadratic equation, it should be solved by rearranging and using the quadratic formula. However, let's use an approach called the ICE approximation to simplify further.
Step 6: Apply the ICE approximation:
Since x is expected to be very small (as it comes from the reactant concentrations), we can neglect it in the denominator. Therefore, we will make the approximation:
Kc = 4x^2 / 0.09 ≈ 53.3
Step 7: Solve for x:
4x^2 ≈ 0.09 * 53.3
4x^2 ≈ 4.797
x^2 ≈ 1.199
x ≈ √1.199 ≈ 1.095
Step 8: Calculate the equilibrium concentrations:
[HI] = 2x ≈ 2 * 1.095 ≈ 2.19 M
Therefore, the concentration of HI at equilibrium is approximately 2.19 M.