Question

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 = mc^2`.

Answer 1

`DeltaE = -3.892 xx 10^11 "J/mol"`

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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:

`Deltam = m_f - m_i = 4.02387 - 4.02820`

`= -"0.00433 amu"`, or `"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 `"kg/mol"`!

`color(blue)(DeltaE) = Deltamc^2`

`= (-0.00433 xx 10^(-3) "kg/mol")(2.998 xx 10^(8) "m/s")^2`

`= color(blue)(-3.892 xx 10^11 "J/mol")`