Question

Calculate the reaction energy $\mathit{Q}$(in $\mathbf{MeV}$) for the nucleosynthesis reaction ${}_{\mathbf{6}}{}^{\mathbf{12}}\mathbf{C}\mathbf{+}{}_{\mathbf{2}}{}^{\mathbf{4}}\mathit{H}\mathit{e}\mathbf{\to }{}_{\mathbf{8}}{}^{\mathbf{16}}\mathbf{O}$Is this reaction endoergic or exoergic?

Short answer

The reaction energy $O$for nucleon synthesis reaction localid="1667796057861" ${}_6{}^2\mathrm{C}-{}_2\mathrm{He}\to {}_3{}^{13}\mathrm{O}$is $7.161\mathrm{MeV}$, and reaction is exoergic.

Step-by-step solution

Define the reaction energy

The reaction energy is determined by using the expression:

$\mathit{Q}\mathbf{=}\mathbf{[}{\mathbf{M}}_{\mathbf{A}}\mathbf{+}{\mathbf{M}}_{\mathbf{B}}\mathbf{-}{\mathbf{M}}_{\mathbf{C}}\mathbf{-}{\mathbf{M}}_{\mathbf{D}}\mathbf{]}\mathbf{\times }{\mathit{c}}^{\mathbf{2}}$

Where, ${\mathit{M}}_{\mathbf{A}}$and ${\mathit{M}}_{\mathbf{B}}$are the mass of reactants, ${\mathit{M}}_{\mathbf{C}}$and ${\mathit{M}}_{\mathbf{D}}$are the mass of products.

c is the speed of light that and ${\mathit{c}}^{\mathbf{2}}\mathbf{=}\mathbf{931}\mathbf{.}\mathbf{5}\mathit{M}\mathit{e}\mathit{V}\mathbf{/}\mathit{u}$.

Determine the reaction energy

The given reaction is:

${}_6{}^2\mathrm{C}-{}_2{}^{\text{'}}\mathrm{He}\to {}_3{}^{13}\mathrm{O}$

The atomic mass of $\mathrm{He}$is $4.002503u$, the atomic mass of $\mathrm{C}$is $12.0\mathrm{u}$and the atomic mass of O is $15.994915u$

Now,

$\mathrm{Q}=\left[M\left({}_6{}^2\mathrm{C}\right)+M\left({}_2{}^4\mathrm{He}\right)-M\left({}_8{}^6\mathrm{O}\right)\right]$

Substituting the values, and we get,

$\begin{array}{rcl}Q& =& \left[\right(12.0)+(4.002603)-(15.994915\left)\right](931.5)\\ & =& (0.007688)(931.5)\\ & =& 7.161\mathrm{MeV}\\ & & \end{array}$

As the reaction energy Q is positive, so the reaction is exoergic.

Hence, the reaction energy Q for nucleosynthesis reaction localid="1667796103714" ${}_6{}^{12}C+{}_2{}^{4}He\to {}_8{}^{16}O$is localid="1667796107219" $7.161MeV$, and reaction is exoergic.