Question

The quantum yield of the reaction at $330$nm is $0.4$. A sample of acetone absorbs monochromatic radiation at $330nm$ at the rate of $7.2\times {10}^{-3}J{s}^{-1}$ (Given : $N_A=6\times {10}^{23};h=6.6\times {10}^{34}$ in S.I. unit). The rate of formation of $CO\left(mol/s\right)$ is :

• A. $2\times {10}^{-8}$
• B. $8\times {10}^{-8}$
• C. $8\times {10}^{-9}$
• D. none of these

Answer 1

Answer: (C)

$8\times {10}^{-9}$

Energy of one photon $E=\frac{hc}{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{330\times {10}^{-9}}=6\times {10}^{-19}J$
Number of molecules of CO formed per second $\frac{7.2\times {10}^{-3\times 0.4}}{6\times {10}^{-19}}=4.8\times {10}^{15}$
Number of moles of CO formed per second $\frac{4.8\times {10}^{15}}{6.023\times {10}^{23}}=8\times {10}^{-9}mol/s$