# A Level Chemistry 2017 Paper 1 Question 12 Solution - Exclusive

In this exclusive video we want to discuss the suggested solution for A Levels Chemistry (H2 Chemistry) 2017 Paper 1 Question 12.

Let's take a look at this question:

The topic tested in this question is on Kinetics.

The rate equation, which we can determine from experiment, is linked to the mechanism of the reaction.

The order of reaction with respect to a reactant is the number of that reactant involved in the rate determining step or slow step in the mechanism.

So based on this rate equation given, there will be m moles of A and n moles of B in the rate determining step.

So what we have to do is to go through the 4 options and see which one has its rate equation consistent with the mechanism.

## Option A

In the slow step there are 2 NO and 1 H_{2}. So the order of reaction with respect to NO will be 2, and order with respect to H_{2} will be 1.

The rate equation based on the mechanism will be:

rate = k [NO]^{2}[H_{2}]

This is consistent with the option given in the question so it is highly likely to be the answer.

## Option B

In the slow step there is only 1 H_{2}. So the order of reaction with respect to H_{2} will be 1.

The rate equation based on the mechanism will be:

rate = k [H_{2}]

This is not consistent with the option given in the question so it is eliminated.

## Option C

In the slow step there is 1 HBrO and 1 HBr. So the order of reaction with respect to HBrO will be 1, and order with respect to HBr will be 1.

Since HBrO is an intermediate, we need to replace it with the reactants that form this intermediate in the previous fast step.

From the first step, 1 HBrO is formed from 1 HBr and ^{1}/_{2} O_{2}.

Therefore we can replace the HBrO term in the rate equation with HBr order 1 and O_{2} order ^{1}/_{2}.

Finally we can have the rate equation:

rate = k [HBr]^{2}[O_{2}]^{1/2}

This is not consistent with the option given in the question so it is eliminated.

## Option D

In the slow step there is 1 H_{2}O_{2} and 1 IO^{-}. So the order of reaction with respect to H_{2}O_{2} will be 1, and order with respect to IO^{-} will be 1.

Since IO^{-} is an intermediate, we need to replace it with the reactants that form this intermediate in the previous fast step.

From the first step, IO^{-} is formed from 1 H_{2}O_{2} and 1 I^{-}.

Therefore we can replace the IO^{-} term in the rate equation with H_{2}O_{2} order 1 and I^{-} order 1.

Finally we can have the rate equation:

rate = k [H_{2}O_{2}]^{2}[I^{-}]

This is not consistent with the option given in the question so it is eliminated.

So finally we can determine the correct answer for this question has to be option A.

Check out this video for the full solution and detailed calculation!

Topic: Kinetics, Physical Chemistry, A Level Chemistry, Singapore

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