Calculate the reaction energy Q for the reaction $\mathit{p}\mathbf{+}{}_{\mathbf{1}}{}^{\mathbf{3}}\mathit{H}\mathbf{\to }{}_{\mathbf{1}}{}^{\mathbf{2}}\mathit{H}\mathbf{+}{}_{\mathbf{1}}{}^{\mathbf{2}}\mathit{H}$. Is this reaction exoergic or endoergic?

The Reaction Energy Q is defined as; if initial particles A and B interact to produce final particles C and D;

$Q=[M_A+M_B-M_C-M_D]\times 931.5MeV/u$

As Given;

Atomic mass of ${}_1{}^{1}HisM\left({}_1{}^{1}H\right)=1.007825u$.

Atomic mass of ${}_1{}^{2}HisM\left({}_1{}^{2}H\right)=2.014102u$.

Atomic mass of ${}_1{}^{3}HisM\left({}_1{}^{3}H\right)=3.016029u$.

Because the atomic masses of each element in the reaction are not the same as the mass of the nuclei, the reaction equation must be corrected.

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

The number of electrons on both sides of the equation is equal.

Putting the numbers together, the reaction energy is:

$$\begin{gathered} Q=[3.016029u-2\times 2.014102u+1.007825u]\times 931.5MeV \\ =[-0.004330]\times 931.5MeV \\ =-4.033MeV \end{gathered}$$

The reaction is endoergic because the reaction energy is negative.

Hence, the reaction energy Q = - 4.033 MeV and it is endoergic.