Chapter 6, Problem 6E.3E

Interpretation Introduction

Interpretation:

From the given data the relaxation time corresponding for the return to equilibrium after a temperature jump has to be calculated.

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. 

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

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

On rearranging, $\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}={\text{K}}_{\text{c}}$

Where,

${\text{k}}_f$ is the rate constant of the forward reaction.

${\text{k}}_r$ is the rate constant of the reverse reaction.

$K_c$ is the equilibrium constant.

Half-life period:

Half-life is defined as the time taken for amount of the sample to reduce to half of its starting value.  The symbol is ${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$.

  $t_{\frac{1}{2}}=\frac{\ln \left(2\right)}{k}$

Half-life for a first order reaction is, $t_{\frac{1}{2}}=\frac{0.693}{k}$

First order reaction $-kt=\ln \left(\frac{\left[A\right]}{\left[A_0\right]}\right)$

$\left[A\right]$ is the concentration of reactant A at time t ${\left[A\right]}_0$ is the  initial concentration of reactant, k is the rate constant.