Chapter 20, Problem 20.128QP

Interpretation Introduction

Interpretation:

• A 215g of Plutonium-239 undergoes decay.  The energy obtained from this decay has to be calculated.
• If an electric energy released from this quantity of Pu-239 is 25% of this value.  The small grams of Zn required for standard voltaic cell to produce same electric energy has to be calculated.

Answer to Problem 20.128QP

• Energy released when 215g of Plutonium-239 undergoes decay is $\text{4}{\text{.559×10}}^{\text{8}}\text{J}$.
• small grams of Zn required for standard voltaic cell to produce same electric energy is $\text{3}{\text{.51×10}}^4\text{g}$

Explanation of Solution

Given

Mass of electron = $\text{0}\text{.000549}\text{amu}$

Temperature = $\text{25°C}$

${}_{\text{94}}^{\text{239}}\text{Pu}\stackrel{}{\to }{}_{\text{92}}^{\text{235}}{\text{U + }}_{\text{2}}^{\text{4}}\text{H}$

Calculate nuclear masses

$\begin{array}{l}{\text{Nuclear mass of }}_{\text{94}}^{\text{239}}\text{Pu = 239}\text{.05216 amu - (94×0}\text{.000549 amu) = 239}\text{.000554 amu }\\ {\text{Nuclear mass of }}_{\text{92}}^{\text{235}}\text{U = 235}\text{.04392 amu - (92×0}\text{.000549 amu) = 234}\text{.993412 amu }\\ {\text{Nuclear mass of }}_{\text{2}}^{\text{4}}\text{He= 4}\text{.00260 amu - (2×0}\text{.000549 amu) = 4}\text{.001502 amu}\end{array}$

$\begin{array}{l}\text{On a mole basis, the mass difference }\Delta \text{m is, }\\ \Delta \text{m = }\left[\text{234}\text{.993412 + 4}\text{.001502 - 239}\text{.000554}\right]\text{ g/mol = -0}\text{.005640 g/mol }\\ \text{= -5}\text{.640}\times {\text{10}}^{\text{-6}}\text{kg/mol}\end{array}$

$\begin{array}{l}\text{The energy released is ,}\\ \Delta \text{E = (}\Delta {\text{m)c}}^{\text{2}}\text{= -5}\text{.640}\times {\text{10}}^{\text{-6}}\text{kg/mol Pu}\times {\text{(2}}^{\text{.998}}\\ \text{= -5}\text{.069}\times {\text{10}}^{\text{11}}{\text{kg·m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \text{= -5}\text{.069}\times {\text{10}}^{\text{11}}\text{J/mol Pu}\end{array}$

$\begin{array}{l}\text{For 215 mg of plutonium-239 the energy released is ,}\\ {\text{ΔE = 215×10}}^{\text{-3}}\text{g Pu×}\frac{\text{1mol Pu}}{\text{239}\text{.05216g Pu}}\text{×}\frac{\text{-5}{\text{.069×10}}^{\text{11}}\text{J}}{\text{1mol Pu}}\\ \text{= -4}{\text{.559×10}}^{\text{8}}\text{J}\end{array}$

$\begin{array}{l}\text{Now calculate 25}\text{.0\% of this energy}\text{. }\\ \text{4}\text{.559}\times {\text{10}}^{\text{8}}\text{J}\times \text{0}\text{.250 = -1}\text{.140}\times {\text{10}}^{\text{8}}\text{J}\end{array}$

$\begin{array}{l}{\text{Next calculate E°for the standard voltaic Zn/Cu}}^{\text{2+}}\text{cell}\text{.}\hfill \\ {\text{ E°}}_{\text{cell}}\text{= 0}\text{.34 V-}\left(\text{-0}\text{.76 V}\right)\text{ = 1}\text{.10 V = 1}\text{.10 V/mol Zn}\hfill \end{array}$

$\begin{array}{l}\text{Now calculate ΔG° }\\ {\text{ΔG°= -nFE°}}_{\text{cell}}\text{= - }\frac{\text{2mol}\text{e}}{\text{1molZn}}\text{×}\frac{\text{9}\text{.65×104C}}{\text{1mol}\text{e}}\text{1}\text{.10 V = -2}{\text{.123×10}}^{\text{5}}\text{J/mol Zn}\end{array}$

$\begin{array}{l}\text{The mass of Zn required to release -1}{\text{.1400×10}}^{\text{8}}\text{J is }\\ \text{ -1}{\text{.140×10}}^{\text{8}}\text{J×}\frac{\text{1mol}\text{Zn}}{\text{-2}{\text{.123×10}}^{\text{5}}\text{J}}\text{×}\frac{\text{65}\text{.38g}\text{Zn}}{\text{1molZn}}\\ \text{= 3}{\text{.51×10}}^{\text{4}}\text{g Zn}\end{array}$

Conclusion

• Energy released when 215g of Plutonium-239 undergoes decay was calculated as $\text{4}{\text{.559×10}}^{\text{8}}\text{J}$.
• small grams of Zn required for standard voltaic cell to produce same electric energy was calculated as $\text{3}{\text{.51×10}}^4\text{g}$