Question

In an experiment, the initial rate of the reaction [s2o82-](m) + [i-](m) is measured for different concentrations. The data is given in the table below:

Experiment [s2o82-](m) [i-](m) initial rate [m/s]
1 0.0400 0.0680 4.80 × 10-4
2 0.0400 0.0340 2.40 × 10-4
3 0.0800 0.0340 4.80 × 10-4

What is the initial rate of the reaction when the concentration of [s2o82-] is 0.0800 M and the concentration of [i-] is 0.0680 M?

Answer 1

To find the initial rate of the reaction, we look at how changes in concentration affect the rate. The rate is directly proportional to the concentration of both reactants based on provided experimental data. When the concentrations are both 0.0800 M for S2O82- and 0.0680 M for I-, the initial rate is predicted to be 9.60 × 10^-4 M/s.

Step-by-step explanation:

To determine the initial rate of the reaction when the concentration of [S2O82-] is 0.0800 M and the concentration of [I-] is 0.0680 M, we analyze the given data to understand how the initial rates change with changes in concentration.

From Experiment 1 to Experiment 2, when the concentration of I- is halved, the initial rate also halves (from 4.80 × 10-4 to 2.40 × 10-4 M/s), indicating that the rate is directly proportional to the concentration of I-. Experiment 3 compared with Experiment 2 shows that when the concentration of S2O82- is doubled, the initial rate also doubles (from 2.40 × 10-4 to 4.80 × 10-4 M/s), signifying a direct proportionality between the rate and the concentration of S2O82-.

Based on this proportional relationship, if we double the concentration of [I-] to 0.0680 M in the context of Experiment 3's concentration of S2O82-, we could predict that the initial rate would also double. Thus, doubling the initial rate from Experiment 3 that has an initial rate of 4.80 × 10-4 M/s would yield an initial rate of 9.60 × 10-4 M/s for the given concentrations.