For CH4+Cl2⟶hv product, rate of reaction at t=0 is given by rate =-ddtCH4=2×10-4 M-min-1, then rate of the reaction at t=20 min is (in M-min-1 )

rate = − d dt CH 4 = 4 × 10 − 4

rate = − d dt CH 4 = 4 × 10 − 4

rate = − d dt CH 4 = 10 − 4

rate = − d dt CH 4 = 4 × 10 − 4

rate = − d dt Cl 2 = 2 × 10 − 4

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For ${CH}_{4} + {Cl}_{2} \overset{h v}{⟶}$ product, rate of reaction at $t = 0$ is given by rate $= \frac{- d}{dt} [{CH}_{4}] = 2 \times 10^{- 4}$ $M - \min^{- 1}$ , then rate of the reaction at $t = 20$ $\min$ is (in $M - \min^{- 1}$ )

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$rate = \frac{- d}{dt} [{CH}_{4}] = 10^{- 4}$

$rate = \frac{- d}{dt} [{CH}_{4}] = 4 \times 10^{- 4}$

$rate = \frac{- d}{dt} [{Cl}_{2}] = 2 \times 10^{- 4}$

$rate = \frac{- d}{dt} [{Cl}_{2}] = zero$

answer is B.

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The given reaction is:

${CH}_{4} + {Cl}_{2} \overset{h v}{⟶}$ product

It is a photochemical reaction so, it is a zero-order reaction.

The rate of the reaction will be written as:

Rate = $\frac{- d}{dt} [{Cl}_{2}] = \frac{- d}{dt} [{CH}_{4}] = 2 \times 10^{- 4}$ $M - \min^{- 1}$

Therefore, the correct option is (B).

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