Chapter 5, Problem 65P

Interpretation Introduction

Interpretation:

Using the given data of rate constant at different temperature, activation energy,$\Delta G^0$,$\Delta H^0$ and $\Delta S^0$ should be determined for the reaction at ${30}^{0}C$.

Concept introduction:

• Arrhenius equation used to find the rate constant by showing the fraction of collisions with effective orientation and represents the fraction of collisions with sufficient energy to produce a reaction.  Also this reaction gives the quantitative relation of temperature with the rate constant of reaction.

$\boxed{k=A{e}^{\frac{{}^{-E_a}}{RT}}}$

k is the rate constant, A is the frequency factor, $E_a$ is the activation energy

R is the gas constant and T is the temperature.

If rate constant at several temperature is known, using Arrhenius equation activation energy,$\Delta G^0$,$\Delta H^0$ and $\Delta S^0$ can be calculated.

• Activation energy $\left(E_a\right)$ can be obtained using Arrhenius equation from the slope of a plot of ln k versus because,

  ${\text{lnk}}_{\text{2}}{\text{-lnk}}_{\text{1}}=\frac{{\text{-E}}_{\text{a}}}{\text{R}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)}$

• $\Delta H^0=E_a-RT$
• ${\text{-ΔG}}^{\text{0}}\text{=}\text{RTln}\frac{\text{kh}}{{\text{Tk}}_{\text{B}}}\boxed{\begin{array}{l}\text{h}\text{is}\text{the}\text{Plank's}\text{constant}\text{=}\text{1}\text{.58}{\text{×10}}^{\text{-31}}\text{kcal}\text{s}\\ {\text{k}}_{\text{B}}\text{is}\text{the}\text{Boltzmann}\text{constant}\text{=}\text{3}\text{.30}{\text{×10}}^{\text{-19}}\text{kcal}{\text{K}}^{\text{-1}}\end{array}}$
• $\Delta S^0=\frac{\left(\Delta H^0-\Delta G^0\right)}{T}$