How do I solve this type of question

Question

The reaction represented by the balanced chemical equation was studied at 25 oC. The initial rate of reaction was determined by measuring the rates of change in the concentrations of I− (aq) and S2O82− (aq) near the beginning of the reaction (Table 1).

Table 1 Initial Rates from Runs at 25 oC of the Reaction represented by

2 I−(aq) + S2O82− (aq) → I2 (aq) + 2 SO42−(aq)

Trial [I−]0(mol/L) [S2O82−]0(mol/L) Initial Rate(mol/L s)

1 0.080 0.040 12.5 ×10−6

2 0.040 0.040 6.25 ×10−6

3 0.080 0.020 6.25 ×10−6

4 0.032 0.040 5.00×10−6

a) What is the rate law equation for this reaction?

b) What will the rate of reaction be when [I− ] =0.060 mol/L [[S2O82−] = 0.030 mol/L?

Answer 1

Compare trial 2 to trial 1. [S₂O₈^2-] remains constant at 0.04 M but [I^-] doubles from 0.04 to 0.08 M. The rate also doubles from 6.25x10^-6 to 12.5x10^-6. This tells us that the reaction is FIRST ORDER in I^-

Now, compare trial 3 to trial 1. [I^-] remains constant but [S₂O₈^2-] doubles from 0.02 to 0.04 M. The rate also doubles from 6.25x10^-6 to 12.5x10^-6. This tells us that the reaction is FIRST ORDER in S₂O₈^2-.

a) We can now write the rate law as : Rate = k[I^-][S₂O₈^2-]

b) Solve for the rate constant by using the rate and concentrations from any trial above:

12.5x10^-6 M/s = k[0.08 M][0.04 M]

k = 0.00391 M^-1s^-1

And use this to solve for rate of reaction be when [I⁻ ] =0.060 mol/L & [[S₂O₈²⁻] = 0.030 mol/L

Rate = 0.00391 M^-1s^-1 (0.060 M)(0.030 M) = 7.04x10^-6 M/s