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# A2: Orders of reactions Watch

1. 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.
2. (Original post by 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
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.
3. (Original post by 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.
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 t1/2 = 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)
4. (Original post by 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 t1/2 = 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.
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.
5. (Original post by 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 t1/2 = 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)
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.
Thank you both for replying. So if there is a concentration-time graph with equal half lives does that show that the overall order is first? I always assumed it was only referring to the one reactant on the graph's y-axis?

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