For H2+Cl2→ hษ 2HCl, rate of reaction at t=0 is given by rate =-ddtH2=6×10-3M-min-1, then rate of the reaction at t=30 min is

− d dt [ Cl 2 ] = 6 × 10 − 4 M − min − 1

− d dt [ Cl 2 ] = 6 × 10 − 4 M − min − 1

1 2 d dt [ HCl ] = 12 × 10 − 3 M − min − 1

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For $H_{2} + {Cl}_{2} \overset{hษ}{\to} 2 HCl$ , rate of reaction at $t = 0$ is given by rate $= - \frac{d}{dt} [H_{2}] = 6 \times 10^{- 3} M - \min^{- 1}$ , then rate of the reaction at $t = 30 \min$ is

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$\frac{- d}{dt} [{Cl}_{2}] = 6 \times 10^{- 4} M - \min^{- 1}$

Zero

$\frac{1}{2} \frac{d}{dt} [HCl] = 12 \times 10^{- 3} M - min^{- 1}$

answer is A.

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The given reaction is:

$H_{2} + {Cl}_{2} \overset{hv}{\to} 2 HCl$

It is a photochemical reaction, due to which it is a zero-order reaction.

The rate law for the reaction can be written as:

$= - \frac{d}{dt} [H_{2}] = - \frac{d}{dt} [{Cl}_{2}] = \frac{1}{2} \frac{d}{dt} [HCl]$

The given value is $= - \frac{d}{dt} [H_{2}] = 6 \times 10^{- 3} M - \min^{- 1}$

So, for Chlorine it will be:

$= - \frac{d}{dt} [{Cl}_{2}] = 6 \times 10^{- 3} M - \min^{- 1}$

For Ammonia it will be:

$= \frac{1}{2} \frac{d}{dt} [HCl] = \frac{6 \times 10^{- 3} M - \min^{- 1}}{2} = 3 \times 10^{- 3} M - \min^{- 1}$

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