Using the Equilibrium Law

We can use the Equilibrium Law to help us understand what happens when the concentration of a component of an equilibrium mixture is changed.

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Think about the effect of increasing the concentration of HI(g) in an equilibrium mixture of HI(g), H2(g) and I2(g):

2HI(g)  H2(g) + I2(g)

Suppose the original equilibrium concentrations are:

[HI(g)] = 0.07 mol dm–3

[H2(g)] = 0.01 mol dm–3

[I2(g)] = 0.01 mol dm–3

\inline \dpi{100} K_{\textup{c}}= \frac{[\textup{HI}(\textup{g}))]^{2}}{[\textup{H}_{2}(\textup{g}))][\textup{I}_{2}(\textup{g}))]}  \inline \dpi{100} = \frac{(0.07)^{2}}{(0.01))(0.01)}= 49

The concentration of H2(g) is suddenly doubled to become 0.02 mol dm–3.

How will the system respond to this change?

Equilibrium will move to the left to minimise the effect of the change.

What will happen to the concentration of HI(g)?

The concentration will increase.

What will happen to the concentration of I2(g)?

The concentration will decrease.

When equilibrium is re-established the concentration of HI is found to be 0.076 mol dm–3.

By how much has the concentration of HI(g) increased in achieving the new equilibrium?

0.076 – 0.07 = 0.006 mol dm–3

By how much have the concentrations of H2(g) and I2(g) decreased in achieving the new equilibrium?

0.003 mol dm–3

What will be the concentration of H2(g) when equilibrium is re-established?

0.02 – 0.003 = 0.017 mol dm–3

What will be the concentration of I2(g) when equilibrium is re-established?

0.01 – 0.03 = 0.07 mol dm–3

What is the value of Kc that you can calculate using the new values of the equilibrium concentrations?

49

 
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