Question

Suppose that in a given kinetic experiment, the blue color appears in 14.5 seconds when the initial concentrations are [Na, S₂O₃]. =

8.79×10M,
[C]. = 4.58×10M, [H₂O₂]. = 5.10×10M. What is the rate of change of [ls"") in reaction 1?

a. −1/4.5
b. 1/14.5
c. -1/5.10
d. 1/5.10

Answer 1

The rate of change of substance B in reaction 1 is -4.94×10^-11M/s.

Step-by-step explanation:

The rate of change of substance B in reaction 1 can be determined using the rate expression for the reaction. In this case, the rate expression is given as the change in concentration of A divided by the time. The concentration of A decreases from 8.79×10^-10M to 4.22×10^-10M in 14.5 seconds. Therefore, the rate of change of substance B is:

(4.22×10^-10M - 8.79×10^-10M) / 14.5 seconds = -4.94×10^-11M/s