Chemistry question on rates law

Question

Consider the following reaction of the peroxydisulphate ion (S2O8 2– ) with iodide ion (I– ) in aqueous solution: S2O8 2– (aq) + 3I – (aq) → 2SO4 2– (aq) + I3 – (aq)

At a particular temperature, the rate of this reaction varies with reactant concentrations in the following manner:

Experiment Initial concentration (M) Initial rate ( M .s–1 )

S2O8 2– I –

1 0.038 0.060 1.4 x10–5

2 0.076 0.060 2.8 x10–5

3 0.076 0.030 1.4 x10–5

Determine the rate law for the reaction. Show all your calculations.

Determine the rate constant for the reaction.

If, in a volume of 2.00 dm3 , 0.15 moles of S2O8 2– and 0.060 moles of I – were brought together, what will be rate when 80% of I – has reacted?

Answer 1

The rate law can be written as Rate = k[S₂O₈^2-]^m[I^-]ⁿ where m and n are the order with respect to S₂O₈^2- and I^-, respectively. Our job is to first find m and n, and then find k.

Comparing experiments 3 to 2, we see that [[S₂O₈^2-] remains constant while [I^-] doubles from 0.030 M to 0.060 M. The rate also doubles from 1.4x10^-5 to 2.8x10^-5. This tells us the reaction is FIRST ORDER in I^-.

Thus, n = 1.

Comparing experiments 1 to 2, we see that [I-] remains constant while [[S₂O₈^2-] doubles. The rate also doubles. This tells us the reaction is FIRST ORDER in [S₂O₈^2-]. Thus, m = 1.

Rate Law: Rate = k[[S₂O₈^2-][I^-]

Determining rate constant (k): Simply use the above rate law and substitute data from any experiment. We'll use experiment 1.

Rate = k[[S₂O₈^2-][I^-]

1.4x10^-5 M/s = k(0.038 M)(0.060 M)

1.4x10^-5 M/s = 0.00228 M² k

k = 6.14x10^-3 M^-1s^-1

If, in a volume of 2.00 dm3 , 0.15 moles of S₂O₈^2- and 0.060 moles of I^- were brought together, what will be rate when 80% of I^- has reacted?

Initial [S₂O₈^2-] = 0.15 mol/2 L = 0.075 M

Initial [I-] = 0.060 mol/2 L = 0.030 M

If 80% of I^- reacts, then ~27% of S₂O₈^2- will also have reacted since mol ratio is 3:1 for I- to S₂O₈^2-.

This leaves ~73% of the S₂O₈^2- and 20% of the I^-:

[S₂O₈^2-] = 0.73 x 0.075 M = 0.0548 M

[I^-] = 0.20 x 0.030 M = 0.006 M

Rate = k[S₂O₈^2-][I^-]

Rate = 6.14x10^-3 M^-1s^-1 (0.0548 M)(0.006 M)

Rate = 2.1x10^-6 M/s