Chapter 5, Problem 5.23P

Interpretation Introduction

(a)

Interpretation:

Activation energy is to be calculated.

Concept Introduction:

To calculate the activation energy for the diffusion coefficient at temperature ${727}^{\text{o}}C$ the equation used is as follows:

  $D=D_0\exp \left(-\frac{Q}{RT}\right)\mathrm{..........}\left(1\right)$

Here, $D$ is the diffusion coefficient of $C{r}^{-3}{\text{ in Cr}}_2{O}_3,Q$ is the activation energy, gas constant is $R$ and the absolute temperature is $T$.

Answer to Problem 5.23P

The activation energy is $59230cal/mol$

Explanation of Solution

Calculate the equation for the diffusion coefficient at temperature ${727}^{\text{o}}C$ is follows.

  $D=D_0\exp \left(-\frac{Q}{RT}\right)\mathrm{..........}\left(1\right)$

Here, $D$ is the diffusion coefficient of $C{r}^{-3}{\text{ in Cr}}_2{O}_3,Q$ is the activation energy, gas constant is $R$ and the absolute temperature is $T$.

Substitute $6\times {10}^{-15}c{m}^2/s\text{ for }D\left(1.987cal/mol.K\right)\text{ for }R$ ,

Convert the temperature is in Celsius to Kelvin scale or the absolute temperature scale as follows.

  $T_k=273+T_c\mathrm{.........}\left(2\right)$

Here, temperature in Kelvin is $T_k$ temperature in Celsius is $T_c$.

Substitute ${727}^{0}C$ for the value of $T_c$ in equation $\left(2\right)$.

  $\begin{array}{c}{T}_k=273+T_c\\ =273+723\\ =1000K\end{array}$

Arrhenius equation uses the value of absolute temperature.

Hence $T$ in Arrhenius equation is the same as the $T_k$.

Substitute the above value in equaion $\left(1\right)$

  $\begin{array}{c}6\times {10}^{-15}=D_0\exp \left(-\frac{Q}{RT}\right)\\ =D_0\exp \left(-\frac{Qcal/mol}{\left(1.987cal/mol.K\right)\times 1000K}\right)\mathrm{..........}\left(3\right)\\ =D_0\exp \left(-\frac{Q}{1987}\right)c{m}^2/s\end{array}$

Calculate the equation for the diffusion coefficient at temperature ${1400}^{\text{o}}C$ is follows.

  $D=D_0\exp \left(-\frac{Q}{RT}\right)$

Here, $D$ is the diffusion coefficient of $C{r}^{-3}{\text{ in Cr}}_2{O}_3,Q$ is the activation energy, gas constant is $R$ and the absolute temperature is $T$.

Substitute $6\times {10}^{-15}c{m}^2/s\text{ for }D\left(1.987cal/mol.K\right)\text{ for }R$ ,

Convert the temperature is in Celsius to Kelvin scale or the absolute temperature scale as follows.

  $T_k=273+T_c$

Here, temperature in Kelvin is $T_k$ temperature in Celsius is $T_c$.

Substitute ${1400}^{0}C$ for the value of $T_c$ in equation $\left(2\right)$.

  $\begin{array}{c}{T}_k=273+T_c\\ =273+1400\\ =1673K\end{array}$

Arrhenius equation uses the value of absolute temperature.

Hence $T$ in Arrhenius equation is the same as the $T_k$.

Substitute the above value in equaion $\left(1\right)$ to obtain.

  $\begin{array}{c}6\times {10}^{-15}=D_0\exp \left(-\frac{Q}{RT}\right)\\ =D_0\exp \left(-\frac{Qcal/mol}{\left(1.987cal/mol.K\right)\times 1673K}\right)c{m}^2/s\\ =D_0\exp \left(-\frac{Qcal/mol}{3324.251}\right)\\ =D_0\exp \left(-\frac{Q}{1987}\right)c{m}^2/s\mathrm{..........}\left(4\right)\end{array}$

Calculate the activation energy.

Divide equaiton $\left(3\right)$ by equation $\left(4\right)$.

  $\begin{array}{c}\frac{6\times {10}^{-15}c{m}^2/s}{1\times {10}^{-9}c{m}^2/s}=\frac{D_0\exp \left(-\frac{Q}{1987}\right)c{m}^2/s}{D_0\exp \left(-\frac{Q}{3324.251}\right)c{m}^2/s}\\ 6\times {10}^{-6}=\exp \left[-Q\left(0.000503-0.0003\right)\right]\\ 6\times {10}^{-6}=\exp \left[-0.000203Q\right]\\ \ln \left(6\times {10}^{-6}\right)=\left[-0.000203Q\right]\\ -12.024=-0.000203Q\\ Q=59230cal/mol\end{array}$

Therefore, the activation energy is $59230cal/mol$.

Interpretation Introduction

(b)

Interpretation:

The constant $D_0$ is to be calculated.

Concept Introduction:

Calculate the value of constant $D$.

Substitute the value of the activation energy $Q$ in diffusion constant equation $\left(1\right)$.

  $D=D_0\exp \left(-\frac{Q}{RT}\right)$

Answer to Problem 5.23P

The constant $D_o$ is $0.055c{m}^2/s$..

Explanation of Solution

Calculate the value of constant $D$.

Substitute the value of the activation energy $Q$ in diffusion constant equation $\left(1\right)$.

  $D=D_0\exp \left(-\frac{Q}{RT}\right)$

Substitute $1\times {10}^{-9}c{m}^2/s\text{ for }D,59230cal/mol\text{ for }Q,1673K\text{for}T\text{and 1}\text{.987cal/mol}\text{.}K$ for $R$.

  $\begin{array}{c}1\times {10}^{-9}c{m}^2/s=D_0\exp \left(-\frac{Q}{RT}\right)c{m}^2/s\\ =D_0\exp \left(-\frac{59230cal/mol}{\left(1.987cal/mol.K\right)\times 1673K}\right)c{m}^2/s\\ =D_0\exp \left(-17.818\right)\\ =1.828\times {10}^{-8}{D}_0\end{array}$

  $D_0=0.055c{m}^2/s$

Hence, the constant $D_o\text{ is }0.055c{m}^2/s$