Chapter 10.14, Problem 85AAP

To determine

The activation energy for elastomer relaxation process using an Arrhenius-type rate equation.

Answer to Problem 85AAP

The activation energy for elastomer relaxation process using an Arrhenius-type rate equation is $\underset{\_}{\text{37}\text{.65}\text{kJ}/\text{mol}}$.

Explanation of Solution

Express the first relaxation time.

    ${\tau }_1=\frac{-t}{\ln \left(\sigma /{\sigma }_0\right)}$                         (I)

Express the second relaxation time.

    ${\tau }_2=\frac{-t}{\ln \left(\sigma /{\sigma }_0\right)}$                         (II)

Express the Arrhenius-type rate equation.

For first,

    $\frac{1}{{\tau }_1}=C{e}^{-Q/RT}$                            (III)

For second,

    $\frac{1}{{\tau }_2}=C{e}^{-Q/RT}$                           (IV)

Dividing the Equation (IV) and (III).

    $\frac{{\tau }_2}{{\tau }_1}=\exp \left[\frac{-Q}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right]$                (V)

Conclusion:

Unit conversion of temperature from Celsius to Kelvin.

  $\begin{array}{c}{T}_1=27°\text{C}\\ =27°\text{C}+273\\ =300\text{K}\end{array}$

  $\begin{array}{c}{T}_2=50°\text{C}\\ =50°\text{C}+273\\ =323\text{K}\end{array}$

The average molecular mass of the butadiene mer is $54\text{g}/\text{mol}$.

The average molecular mass of the Sulphur is $32\text{g}/\text{mol}$.

Substitute 25 days for $t$, 750 psi for $\sigma$, and 1000 psi for ${\sigma }_0$ in Equation (I).

    $\begin{array}{c}{\tau }_1=\frac{-\left(25\text{days}\right)}{\ln \left(750\text{psi}/1000\text{psi}\right)}\\ =\frac{-\left(25\text{days}\right)}{\ln \left(0.75\right)}\\ =\frac{-\left(25\text{days}\right)}{-0.28768}\\ =86.90\text{days}\end{array}$

Substitute 30 days for $t$, 400 psi for $\sigma$, and 1100 psi for ${\sigma }_0$ in Equation (II).

    $\begin{array}{c}{\tau }_1=\frac{-\left(30\text{days}\right)}{\ln \left(400\text{psi}/1100\text{psi}\right)}\\ =\frac{-\left(30\text{days}\right)}{\ln \left(0.3636\right)}\\ =\frac{-\left(25\text{days}\right)}{-1.0116}\\ =29.66\text{days}\end{array}$

Substitute $29.66\text{days}$ for ${\tau }_2$, $86.90\text{days}$ for ${\tau }_1$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, 300 K for $T_1$ and 323 K for $T_2$ in Equation (V).

    $\begin{array}{l}\frac{29.66\text{days}}{86.90\text{days}}=\exp \left[\frac{-Q}{\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}\left(\frac{1}{300\text{}\text{K}}-\frac{1}{323\text{K}}\right)\right]\\ Q=\left(\frac{\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}{2.3736\times {10}^{-4}/\text{K}}\right)\ln \left(\frac{29.66\text{days}}{86.90\text{days}}\right)\\ Q=37,653\text{J}/\text{mol}\\ Q=37,653\text{J}/\text{mol}\times \left(\frac{1\text{kJ}/\text{mol}}{1000\text{J}/\text{mol}}\right)\end{array}$

    $Q=37.65\text{kJ}/\text{mol}$

Thus, the activation energy for elastomer relaxation process using an Arrhenius-type rate equation is $\underset{\_}{\text{37}\text{.65}\text{kJ}/\text{mol}}$.