An equilibrium mixture in a 10.0 L flask contains 7.0 mol HI(g) and 1.0 mol each of I2(g) and H2(g). If 2.5 mol H2, 5.0 mol I2 and 2.0 mol HI are added to that equilibrium mixture how many moles of each gas will be present when equilibrium is reestablished?
I worked this for Jon earlier. Post any follow up questions here.
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To find the moles of each gas present when equilibrium is reestablished, we need to use the balanced chemical equation and the concept of equilibrium constants.
The balanced chemical equation for the reaction is:
2 HI(g) ⇌ H2(g) + I2(g)
The equilibrium constant expression (Kc) for this reaction is:
Kc = ([H2][I2]) / [HI]^2
First, let's calculate the initial concentrations of the gases in the equilibrium mixture:
[H2] = 1.0 mol / 10.0 L = 0.1 M
[I2] = 1.0 mol / 10.0 L = 0.1 M
[HI] = 7.0 mol / 10.0 L = 0.7 M
Now, let's calculate the change in moles for each gas when the additional amounts are added:
[H2] will increase by 2.5 mol
[I2] will increase by 5.0 mol
[HI] will increase by 2.0 mol
Since the volume remains constant at 10.0 L, the new concentrations will be:
[H2] = (0.1 M * 10.0 L + 2.5 mol) / 10.0 L = 0.35 M
[I2] = (0.1 M * 10.0 L + 5.0 mol) / 10.0 L = 0.6 M
[HI] = (0.7 M * 10.0 L + 2.0 mol) / 10.0 L = 0.9 M
Now, we can use these concentrations to calculate the new moles of each gas when equilibrium is reestablished:
[H2] = 0.35 M * 10.0 L = 3.5 mol
[I2] = 0.6 M * 10.0 L = 6.0 mol
[HI] = 0.9 M * 10.0 L = 9.0 mol
Therefore, when equilibrium is reestablished, there will be 3.5 moles of H2, 6.0 moles of I2, and 9.0 moles of HI present in the flask.