Consider the reaction,
Let's say the reaction is first order with respect to A, and zero order with respect to B. In other words,
The rate with respect to A is equal to the rate of change of [A], or
Likewise,
Hence,
and
Solving these gives and
Hence,
and
Now, it must be the case that at any point in time, since the change in B in a given time must equal the change in A in that same period. However, the gradient of a graph of [B] against t is constant, whereas the gradient of a graph of [A] against t varies, as given by the two equations above. So the gradients of these two graphs cannot be identical for all values of t.
This makes no sense. Could somebody please point out where I've gone wrong?
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porkstein
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Last edited by porkstein; 27122010 at 21:05. 
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EierVonSatan
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 27122010 23:32
(Original post by porkstein)
I thought that if a reaction was zero order, the rate was constant? 
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 27122010 23:45
You don't express the reaction rate with a rate constant k and equation for each reactant separately; you combine it into one single expression for the whole reaction.
The rate of reaction would simply be
Where is the rate of consumption of species A, and
because 2 molecules of A and 2 molecules of B are consumed in the reaction to produce 1 of C and 2 of D. Note sure whether this notation is used in H.S. level chemistry but the understanding of what's going on is all that matters(Original post by porkstein)
I thought that if a reaction was zero order, the rate was constant?
* at least, so long as the reaction system is within the concentration bounds of zerothorder operation. For example, the adsorption of propanol (P) onto an alumina catalyst allows for its decomposition into propene to occur. The reaction steps are adsorptiondecompositiondesorption. At fairly low propanol concentration, all the catalyst adsorption sites are occupied and the reaction is occuring at its maximum rate. Increasing [P] has no effect on the rate and, as P is the only reactant, the reaction is zeroth order. Of course if you were to start decreasing [P] significantly, the catalyst sites may not be saturated and the rate dependency would change. This example is completely irrelevent but I thought it might be interesting to see how zeroth order kinetics actually come aboutLast edited by gordoxo; 28122010 at 00:03. 
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(Original post by EierVonSatan)
But you said the reaction was first order The rate of loss of A must be equal to the rate of loss of B according to your chemical equationLast edited by porkstein; 28122010 at 00:18. 
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 28122010 00:07
(Original post by gordoxo)
You don't express the reaction rate with a rate constant k and equation for each reactant separately; you combine it into one single expression for the whole reaction.
The rate of reaction would simply be
Where is the rate of consumption of species A, and
because 2 molecules of A and 2 molecules of B are consumed in the reaction to produce 1 of C and 2 of D. Note sure whether this notation is used in H.S. level chemistry but the understanding of what's going on is all that matters
EDIT: Yes, that makes sense. The rate of reaction is said to depend on the species involved in the slow step of the mechanism, but this is only really valid as long as the slow step is considerably slower than the other steps. Does that sound about right?Last edited by porkstein; 28122010 at 00:12. 
charco
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(Original post by porkstein)
I see, thanks a lot. So what exactly does the rate equation represent? The rate of change of concentration of what? Presumably any of the reactants/products, depending on the value of the rate constant. But I thought rate equations were given with the constant assuming a specific value. In this case, surely the equation is only accurate for one of the species in the chemical equation, and if that is so, how can you tell which one?
EDIT: Yes, that makes sense. The rate of reaction is said to depend on the species involved in the slow step of the mechanism, but this is only really valid as long as the slow step is considerably slower than the other steps. Does that sound about right?
The RDS is assumed to be a step which is slower than any other steps and usually considerably slower. 
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 28122010 00:34
(Original post by porkstein)
I see, thanks a lot. So what exactly does the rate equation represent? The rate of change of concentration of what? Presumably any of the reactants/products, depending on the value of the rate constant. But I thought rate equations were given with the constant assuming a specific value. In this case, surely the equation is only accurate for one of the species in the chemical equation, and if that is so, how can you tell which one?
My memory begins to break down a bit when it comes to how rate equations are defined. I /think/ what generally happens is it is defined using the rate of change of concentration of a chosen main reactant (often the limiting reactant  the one you run out of first). You can then get the rate of change of conc. of all other species by scaling using the reaction stoichiometry. For example:
if the rate equation is given as:
then
in other words, the gradient of a graph of concetration against time for species B will be twice as steep than for A  because every time [A] gets consumed, 2[B] is consumed with it. The gradient for product [C] will be equal to [A] but in the opposite direction  increasing instead of decreasing.
I think you are considering the rate of reaction as happening separately for each reactant involved  hence your partial rate equations and realising that the rates for A and B did not match even though they should. Consider this: every time the reaction happens, a set number of reacting molecules are consumed and set number of products created. That means you can't have separate rates for each chemical involved  their rates of change of conc. are all directly related to one another. 
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 28122010 01:53
(Original post by gordoxo)
...Last edited by porkstein; 28122010 at 11:32. 
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 31122010 08:21
(Original post by porkstein)
Right, I believe I understand now. However, I do have one more question. If it doesn't make sense to talk about rates with respect to only one of the species involved, how do we derive equations involving only one species? By that I mean, if we have the rate equation , how can we graph [B] against t, or d[A]/dt against [A], for instance? You can see from my original post that I have tried to derive conctime graphs by integrating these 'partial rate equations', but surely you cannot do such a thing when your rate equation involves one than one variable? Or are these graphs only really feasible when the rate equation depends on the concentration of only one species?
The problem with obtaining an analytical solution is, as you've noticed, the interdependencies between the variables. You cannot describe [A] solely as a function of time, as [B] also needs to be considered. You have a system of partial differential equations. The problem with PDEs is their analytical solutions are difficult to obtain (analytical as in mathematical equations). The examples used here were simple and there are methods of tackling them (e.g. look at twothirds of the way down here. Second order equation, case 2. Looks horrible eh?) but for the more complex systems I mentioned, getting an analytical solution is frequently impossible.
Numerical integration is used instead. You take your initial conditions, for our example let:
at with the reaction A+2B>C. For simplicity let
Then consider an "infinitely" small timestep. Let's say our example reaction is known to be fairly quick, finishing within ~30 seconds. We might take a timestep of 0.1 seconds.
So in our first 0.1 seconds, we can calculate our rates of change:
and we know from the stoichiometry that
Then you calculate the change in your variables in this first timestep
Concentration of A at t=0.1
Similarly,
We would then calculate our new rates of change (at t=0.1s):
and so on again for B and C rates of change.
then you calculate the concentrations at t=0.2s
You keep doing this for as long as you need to reach a certain minmum concentration or some other condition, and graph all your points through against time. Plotting all your results will give three curves, showing A and B exponentially decreasing to zero and C exponentially increasing to 1M.
You'll notice the calculations are essentially the same equations using new values over multiple iterations. This lends itself very well to spreadsheets or coding (MATLAB rules for anything like this). The advantage of numerical integration is you can use it to get the concentrations profiles for really complicated systems. A disadvantage is that you have to make sure you have a suitably small timestep value  notice how in the example the rate of reaction nearly halves between 0 seconds and 0.1 seconds? Yeah, that means the figures I randomly picked cause the reaction to end much faster and it requires a smaller timestep to avoid large error.
Of course, the rate equation you gave in the OP is just an ODE (because there is no [B] term in the rate equation) and can be solved as normal (analytically) to get the graph you want.
It's 7am, please let me know if anything I've written doesn't make sense or could do with more explanation I'd very much recommend creating a quick spreadsheet using the formulae I mentioned above to perform the numerical integration yourself if you are interested  copying cell formulae down rows means you have to do very little actual work. Let me know how you get on  if you don't know how to use cell formulae in excel or run into problems I'll show you what I mean.
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