Chapter 11.13, Problem 101AAP

To determine

The value for the activation energy in the temperature region.

Answer to Problem 101AAP

The value for the activation energy in the temperature region is $428\text{kJ}/\text{mol}$.

Explanation of Solution

Write the equation involving the viscosity to temperature for a viscous flow in the glass.

  $\eta ={\eta }_0{e}^{Q/RT}$                                                                                                            (I)

Here, viscosity of the glass is $\eta$, pre-exponential constant is ${\eta }_0$,molar activation energy for viscous flow is $Q$, universal gas constant is $R$ and absolute temperature is $T$.

For the strain point, write the equation involving the viscosity to temperature for a viscous flow in the glass.

  ${\eta }_{stp}={\eta }_0{e}^{Q/R{T}_{stp}}$                                                                                                      (II)

Here, viscosity of the glass at strain point is ${\eta }_{stp}$ and strain point temperature is $T_{stp}$.

For the softening point, write the equation involving the viscosity to temperature for a viscous flow in the glass.

  ${\eta }_{sp}={\eta }_0{e}^{Q/R{T}_{sp}}$                                                                                                      (III)

Here, viscosity of the glass at softening point is ${\eta }_{sp}$ and softening point temperature is $T_{sp}$.

Conclusion:

Convert the unit of strain point temperature to absolute value.

 $\begin{array}{c}{T}_{stp}=500°\text{C}\\ =500+273\text{K}\\ =773\text{K}\end{array}$

Convert the unit of softening point temperature to absolute value.

 $\begin{array}{c}{T}_{sp}=700°\text{C}\\ =700+273\text{K}\\ =973\text{K}\end{array}$

Divide Equation (II) by Equation (III).

  $\begin{array}{c}\frac{{\eta }_{stp}}{{\eta }_{sp}}=\frac{{\eta }_0{e}^{Q/R{T}_{stp}}}{{\eta }_0{e}^{Q/R{T}_{sp}}}\\ =\frac{e^{Q/R{T}_{stp}}}{e^{Q/R{T}_{sp}}}\\ =\exp \left[\frac{Q}{R}\left(\frac{1}{T_{stp}}-\frac{1}{T_{sp}}\right)\right]\end{array}$                                                                                  (IV)

Substitute ${10}^{14.2}\text{P}$ for ${\eta }_{stp}$, ${10}^{7.5}\text{P}$ for ${\eta }_{sp}$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $773\text{K}$ for $T_{stp}$ and $973\text{K}$ for $T_{sp}$ in Equation (IV).

 $\begin{array}{l}\frac{{10}^{14.2}\text{P}}{{10}^{7.5}\text{P}}=\exp \left[\frac{Q}{8.314\text{J}/\text{mol}\cdot \text{K}}\left(\frac{1}{773\text{K}}-\frac{1}{973\text{K}}\right)\right]\\ {10}^{6.7}=\exp \left(\left(3.605\times {10}^{-5}\right)Q\right)\\ \ln {10}^{6.7}=\left(3.605\times {10}^{-5}\right)Q\end{array}$

 $\begin{array}{c}Q=4.28\times {10}^5\text{J}/\text{mol}\\ =4.28\text{kJ}/\text{mol}\end{array}$

Thus, the value for the activation energy in the temperature region is $428\text{kJ}/\text{mol}$.