Question

The mass of a hydrogen atom (${}_1^1\mathrm{H}$) is$1.007825$amu; that of a tritium atom (${}_1^3\mathrm{H}$) is$3.01605$amu; and that of an$\alpha$particle is$4.00150$amu. How much energy in kilojoules per mole of${}_4^2\mathrm{H}\mathrm{e}$produced is released by the following fusion reaction:${}_1^1{\mathrm{H}+}_1^3\mathrm{H}{\to }_4^2\mathrm{H}\mathrm{e}$.

Short answer

The energy obtained is:$\mathrm{E}=2\times 10^9\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$.

Step-by-step solution

Step 1: Nuclear Chemistry

Nuclear chemistry is a branch of chemistry that studies radioactivity, nuclear processes, and atomic nuclei alterations such as nuclear transmutation and nuclear characteristics.

Evaluating the energy

We can get energy by following a simple formula:

$\mathrm{E}=\mathrm{m}\times {\mathrm{c}}^2$

But first, we must understand the response that takes place:

${}_1^1{\mathrm{H}+}_1^3\mathrm{H}{\to }_4^2\mathrm{H}\mathrm{e}$

The total mass then will be:

$\begin{array}{c}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}=\mathrm{m}(\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s})-\mathrm{m}(\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s})\\ \mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}=(1.007285\mathrm{a}\mathrm{m}\mathrm{u}+3.01605\mathrm{a}\mathrm{m}\mathrm{u})-4.00150\mathrm{a}\mathrm{m}\mathrm{u}\\ \mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}=0.022375\mathrm{a}\mathrm{m}\mathrm{u}\end{array}$

Then, converting the value of amu in kg is:

$\begin{array}{c}1\mathrm{a}\mathrm{m}\mathrm{u}=1.66\times 10^{-27}\mathrm{k}\mathrm{g}\\ \mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}=0.022375\mathrm{a}\mathrm{m}\mathrm{u}\times 1.66\times 10^{-27}\mathrm{k}\mathrm{g}\times \mathrm{a}\mathrm{m}{\mathrm{u}}^{-1}\\ =3.715\times 10^{-29}\mathrm{k}\mathrm{g}\end{array}$

The energy is then evaluated as:

$\begin{array}{c}\mathrm{E}=\mathrm{m}\times {\mathrm{c}}^2\\ \mathrm{E}=3.715\times 10^{-29}\mathrm{k}\mathrm{g}\times {\left(2.99\times 10^8\mathrm{m}{\mathrm{s}}^{-}1\right)}^2\\ \mathrm{E}=3.322\times 10^{-12}\mathrm{J}\\ \mathrm{E}=3.322\times 10^{-15}\mathrm{k}\mathrm{J}\end{array}$

However, that is energy per nucleus. Multiply it by the Avogadro number to find the energy per mole:

$\begin{array}{c}\mathrm{E}=3.322\times 10^{-15}\mathrm{k}\mathrm{J}\times \mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}{\mathrm{s}}^{-1}\times 6.022\times 10^{23}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s}\times \mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}\\ \mathrm{E}=2\times 10^9\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}\end{array}$

Therefore, the energy is: $\mathrm{E}=2\times 10^9\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$.