For photochemical dimerisation of $C_{12} H_{14}$ to obtain $C_{24} H_{28}$ , the proposed rate law is
$\frac{d [C_{24} H_{28}]}{d t} = \frac{K_{1} K_{3} I_{a b s} [C_{12} H_{14}]}{K_{2} + K_{3} [C_{12} H_{14}]}$
where, $I_{a b s}$ = Intensity of absorbed light.
What will be the value of rate of formation of $C_{24} H_{28}$ if the reaction follows zero order kinetics?

K_{1}

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K_{1} I_{a b s}

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\frac{K_{1} K_{3} I_{a b s}}{K_{2}}

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\frac{K_{1} I_{a b s}}{K_{2}}

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Solution

The correct option is B $K_{1} I_{a b s}$
$\frac{d [C_{24} H_{28}]}{d t} = \frac{K_{1} K_{3} I_{a b s} [C_{12} H_{14}]}{K_{2} + K_{3} [C_{12} H_{14}]} = \frac{K_{1} K_{3} [I]_{a b s o r b}}{K_{2} + K_{3}} = K_{1} [I]_{a b s o r b}$
$C_{12} H_{14} \frac{K_{1} K_{3} [I]_{a b s o r b}}{K_{2} + K_{3}} = K_{1} [I]_{a b s o r b} C_{12} H_{14}^{}$
$C_{12} H_{14}^{} + C_{12} H_{14}_{3} C_{24} H_{28}$
[B] as [ $C_{2} H_{14}$ ] in excess

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