# A2: Orders of reactions

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Hi, I was wondering if someone could help me with this question.

I understand part a, that it's first order with respect to OH- because of the equal half lives. However, part B I thought that the fact the half-life doesn't change when doubling the concentration further supports the idea that OH- is first order?

Also it says that the same amount of 2-bromopropane (an excess amount) is used so that's not effecting the rate.

So I understand how they've deduced the rate equation as being first order wrt to 2-bromopropane and 0 order wrt OH-?

Unless there's a mistake on the graph and 2-bromopropane concentration is supposed to be on the y-axis?

Thank you.

I understand part a, that it's first order with respect to OH- because of the equal half lives. However, part B I thought that the fact the half-life doesn't change when doubling the concentration further supports the idea that OH- is first order?

Also it says that the same amount of 2-bromopropane (an excess amount) is used so that's not effecting the rate.

So I understand how they've deduced the rate equation as being first order wrt to 2-bromopropane and 0 order wrt OH-?

Unless there's a mistake on the graph and 2-bromopropane concentration is supposed to be on the y-axis?

Thank you.

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(Original post by

In part a, you deduce that the

Doubling the concentration of hydroxide ions had no effect on the rate (the half life remained constant) -> so it is zero order with respect to hydroxide ions

**EierVonSatan**)In part a, you deduce that the

**overall**reaction order is 1, not with respect to hydroxide ions.Doubling the concentration of hydroxide ions had no effect on the rate (the half life remained constant) -> so it is zero order with respect to hydroxide ions

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(Original post by

How comes it's showing the overall reaction? I thought that conc-time graphs only showed the reaction order with respect to whichever reactant is on the y-axis? Also in my book it says a characteristic of 1st order reactants is that they have a constant half-life.

**LeanneAmyx**)How comes it's showing the overall reaction? I thought that conc-time graphs only showed the reaction order with respect to whichever reactant is on the y-axis? Also in my book it says a characteristic of 1st order reactants is that they have a constant half-life.

The graph shows a reaction following apparent first order kinetics.

'normal equation' rate = k[2-bromopropane]

^{a}[OH-]

^{b}

Assuming a large excess (it doesn't state large, but ok) then concentration of the bromoalkane is constant and this absorbed into the rate constant:

new rate equation: rate = k'[OH-]

^{b}

the constant half life gives the overall reaction order as 1 i.e. b = 1. We say that the reaction is pseudo-first order.

Part b, states that the initial concentration of hydroxide ions doesn't affect half life, again this is consistent with a first order reaction since t

_{1/2}= ln 2/k' (i.e. there is no concentration term).

At no point, can we determine

*anything*about the order with respect to 2-bromopropane since using this approximation since its constant throughout

So either, the question is

*deeply*flawed, the excess of bromoalkane is not large and thus you can't get anything from the graph of note or I'm losing my marbles.

(Original post by

**charco**)
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(Original post by

Okay, the more I look at this question, the more confusing it gets Ignore what I said previously.

The graph shows a reaction following apparent first order kinetics.

'normal equation' rate = k[2-bromopropane]

Assuming a large excess (it doesn't state large, but ok) then concentration of the bromoalkane is constant and this absorbed into the rate constant:

new rate equation: rate = k'[OH-]

the constant half life gives the overall reaction order as 1 i.e. b = 1. We say that the reaction is pseudo-first order.

Part b, states that the initial concentration of hydroxide ions doesn't affect half life, again this is consistent with a first order reaction since t

At no point, can we determine

So either, the question is

**EierVonSatan**)Okay, the more I look at this question, the more confusing it gets Ignore what I said previously.

The graph shows a reaction following apparent first order kinetics.

'normal equation' rate = k[2-bromopropane]

^{a}[OH-]^{b}Assuming a large excess (it doesn't state large, but ok) then concentration of the bromoalkane is constant and this absorbed into the rate constant:

new rate equation: rate = k'[OH-]

^{b}the constant half life gives the overall reaction order as 1 i.e. b = 1. We say that the reaction is pseudo-first order.

Part b, states that the initial concentration of hydroxide ions doesn't affect half life, again this is consistent with a first order reaction since t

_{1/2}= ln 2/k' (i.e. there is no concentration term).At no point, can we determine

*anything*about the order with respect to 2-bromopropane since using this approximation since its constant throughoutSo either, the question is

*deeply*flawed, the excess of bromoalkane is not large and thus you can't get anything from the graph of note or I'm losing my marbles.They are supposed to use the information that it is first order to see that it could be due to either the bromoalkane or the hydroxide ions.

The second part shows that it's not hydroxide and therefore must be first order with respect to bromopropane.

They are given the "excess" to show that there is more than enough bromoalkane to allow the hydroxide to react completely, not to suggest that it is removed from the rate equation.

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**EierVonSatan**)

Okay, the more I look at this question, the more confusing it gets Ignore what I said previously.

The graph shows a reaction following apparent first order kinetics.

'normal equation' rate = k[2-bromopropane]

^{a}[OH-]

^{b}

Assuming a large excess (it doesn't state large, but ok) then concentration of the bromoalkane is constant and this absorbed into the rate constant:

new rate equation: rate = k'[OH-]

^{b}

the constant half life gives the overall reaction order as 1 i.e. b = 1. We say that the reaction is pseudo-first order.

Part b, states that the initial concentration of hydroxide ions doesn't affect half life, again this is consistent with a first order reaction since t

_{1/2}= ln 2/k' (i.e. there is no concentration term).

At no point, can we determine

*anything*about the order with respect to 2-bromopropane since using this approximation since its constant throughout

So either, the question is

*deeply*flawed, the excess of bromoalkane is not large and thus you can't get anything from the graph of note or I'm losing my marbles.

(Original post by

My take is that the students are not supposed to consider pseudo-orders.

They are supposed to use the information that it is first order to see that it could be due to either the bromoalkane or the hydroxide ions.

The second part shows that it's not hydroxide and therefore must be first order with respect to bromopropane.

They are given the "excess" to show that there is more than enough bromoalkane to allow the hydroxide to react completely, not to suggest that it is removed from the rate equation.

**charco**)My take is that the students are not supposed to consider pseudo-orders.

They are supposed to use the information that it is first order to see that it could be due to either the bromoalkane or the hydroxide ions.

The second part shows that it's not hydroxide and therefore must be first order with respect to bromopropane.

They are given the "excess" to show that there is more than enough bromoalkane to allow the hydroxide to react completely, not to suggest that it is removed from the rate equation.

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