Chapter 6, Problem 6E.1E

Interpretation Introduction

Interpretation:

In the given experiment the rate constant for the loss of the unreacted substrate from the active site has to be determined.

Concept Introduction:

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$: A system is said to be in equilibrium when all the measurable properties of the system remains unchanged with the time.  Equilibrium constant is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction. 

Consider the reaction where the reactant A is giving product B. 

  $\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

Integrated rate law for second order reactions:

Taking in the example of following reaction,

    $\text{aA}\to \text{products}$

And the reaction follows second order rate law,

Then the relationship between the concentration of $\text{A}$ and time can be mathematically expressed as,

    $\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=kt+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}$

The above expression is called as integrated rate for second order reactions.