Answer
Verified.
Both equations have Q value of 24.7 MeV
Work Step by Step
We can find Q of the provided carbon cycle by adding the Q of all individual reactions.
The Q of the carbon cycle is calculated below
$\begin{aligned} Q_{\text {carbon cycle }} &=Q_{1}+Q_{2}+\cdots+Q_{6} \\ &=(1.95 + 1.19+7.55+7.30+1.73+4.97) \mathrm{MeV} \\ &=24.7 \mathrm{MeV} \end{aligned}$
The Q for proton proton cycle is 26.7 MeV but considering the electron-positron annihilations, we can find the actual Q for comparision
$Q_{p-p}=26.7 \mathrm{MeV}-2(1.02 \mathrm{MeV})=24.7 \mathrm{MeV}$
