Question

An important parameter of a photochemical reaction is the quantum efficiency or quantum yield $\left(\varphi \right)$ which is defined as $\frac{molesofthesubstancereacted}{molesofthephotonsabsorbed}$. Absorption of UV radiation decompose acetone according to the reaction ${(CH}_3{)}_{2}CO$ $\stackrel{hv}{\to }$ $C_2{H}_6$ + CO. The quantum yield of the reaction at 330 nm is 0.4. A sample of acetone absorbs monochromatic radiation at 330 nm at the rate of 7.2 ${10}^{-3}$ ${Js}^{-1}$. The rate of formation of $CO$ (mol/s) is:

(Given : $N_A$ = 6 ${10}^{23}$ ; h = 6.6 ${10}^{-34}$ in S.I. unit).

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

Answer 1

Answer: (C)

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

The energy of 1 photon is
$E=\frac{hc}{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{330\times {10}^{-9}}=6\times {10}^{-19}\text{J}$
The energy of 1 mole of photons is obtained by multiplying the energy of one photon with Avogadro's number.
$E=6.023\times {10}^{23}\times 6\times {10}^{-19}\text{J}=361380\text{J/mol}$
The number of moles of photons absorbed in 1 second is $\frac{7.2\times {10}^{-3}\text{J}}{361380\text{J/mol}}=1.99\times {10}^{-8}mol$
The quantum yield is 0.4
$0.4=\frac{\text{moles of CO formed per second}}{\text{moles of photons absorbed per second}}$

$0.4=\frac{\text{moles of CO formed per second}}{1.99\times {10}^{-8}\text{mol}}$

$\text{moles of CO formed per second}=8\times {10}^{-9}$
Hence, the rate of formation of $CO$ is $8\times {10}^{-9}\text{mol/s}$