Question

Consider the following hypothetical aqueous reaction. A flask is charged with .065 mol of A in a total volume of 100.0 mL. The following data are collected:

Time (min) 0 10 20 30 40
Moles of A 0.065 0.051 0.042 0.036 0.031
This is a second order reaction.
What is the value of the rate constant for the reaction?

Answer 1

Answer: 0.422 M⁻¹s⁻¹

Step-by-step explanation: Reaction Rate is the speed of decomposition of the reactant(s) per unit of time.

A Rate Law relates concentration of reactants, rate reaction and rate constant:

`r=k[A]^(x)[B]^(y)`

where

[A] and [B] are reactants concentration

x and y are reaction order, not related to the stoichiometric coefficients

k is rate constant

r is rate

Before calculating rate constant, first we have to determine reaction order.

In this question, the reactio order is 2. So, the rate law for it is

`-(d[A])/(dt) =k[A]^(2)`

and the integrated formula is

`(1)/([A]) =(1)/([A]_(0)) +kt`

in which

[A]₀ is initial concentration of reactant

Then, using initial concentration at initial time and final concentration at final time:

`(1)/(0.031) =(1)/(0.065) +k(40)`

`40k=(1)/(0.031)-(1)/(0.065)`

`40k=32.26-15.38`

k = 0.422

The rate constant for the reaction is 0.422 M⁻¹.s⁻¹