# For the fission reaction 2""_(1)^(2) "H" -> ""_(1)^(3) "H" + ""_(1)^(1) "H", where the isotopic masses are "2.01410 amu", "3.01605 amu", and "1.00782 amu" respectively, what is the change in energy for this reaction in "J/mol"?

## Hint: first calculate the mass defect, and then use $E = m {c}^{2}$.

Jul 26, 2017

$\Delta E = - 3.892 \times {10}^{11} \text{J/mol}$

You should follow the hint. In this case, we should first find the initial and final masses.

m_(""_1^2 H) = "2.01410 amu", for each deuterium atom on the reactants' side.

m_(""_1^3 H) = "3.01605 amu", for the one tritium atom on the products' side.

m_(""_1^1 H) = "1.00782 amu", for the hydrogen atom on the products' side.

The initial mass is:

${m}_{i} = {\sum}_{k} {m}_{k , i}$

= 2 xx m_(""_1^2 H) = "4.02820 amu"

And the final mass is:

${m}_{f} = {\sum}_{l} {m}_{k , f}$

= m_(""_1^3 H) + m_(""_1^1 H) = "4.02387 amu"

So, the change in mass is:

$\Delta m = {m}_{f} - {m}_{i} = 4.02387 - 4.02820$

$= - \text{0.00433 amu}$, or $\text{g/mol}$.

And that is your mass defect. The change in energy associated with this mass defect is then going to use the change in mass in $\text{kg/mol}$!

$\textcolor{b l u e}{\Delta E} = \Delta m {c}^{2}$

$= {\left(- 0.00433 \times {10}^{- 3} \text{kg/mol")(2.998 xx 10^(8) "m/s}\right)}^{2}$

$= \textcolor{b l u e}{- 3.892 \times {10}^{11} \text{J/mol}}$