Alevel ocr a chem order graphs conc vs time
Can someone explain the concentration time graphs for orders. Like it shows the concentration of the reactants and taht if it doubles , say tis 0 order theres no effect on teh rate. Looking at the 0 order graph I get as time goes it decreases linearly I think thats fine the 1st order graph, why does it ave a curve, is it taht teh rate increases it enables the reversible reaction so it products more of teh reactant, then the 2nd order graph, very steep then pleateaus, so rate is higher when more product then starts to reach an equilibrium?... Can someone explain these please?
Thank you
Thank you
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by LisaPereraCan someone explain the concentration time graphs for orders. Like it shows the concentration of the reactants and taht if it doubles , say tis 0 order theres no effect on teh rate. Looking at the 0 order graph I get as time goes it decreases linearly I think thats fine the 1st order graph, why does it ave a curve, is it taht teh rate increases it enables the reversible reaction so it products more of teh reactant, then the 2nd order graph, very steep then pleateaus, so rate is higher when more product then starts to reach an equilibrium?... Can someone explain these please?
Thank you
Thank you
Let’s assume we have a reaction where some compound A is consumed. The rate equation should take the form:
rate = k [A]^x
If you have done A level maths, you will know that the rate of a reaction can be expressed as something called a derivative- in this case, -d[A]/dt, because [A] decreases as t increases. Hence:
d[A]/dt = -k [A]^x
We can rearrange this slightly to
[A]^-x d[A] = -k dt
Whilst it’s not strictly mathematically correct (since derivatives are not fractions), it does work by coincidence.
Let’s consider what happens if the reaction is zero order (x = 0). Our rearranged equation becomes
d[A] = - k dt (since anything to the power of 0 is 1)
We can carry out an operation called integration on both sides of this expression to obtain
[A] = C - kt
Which if we compare to y = mx + c, we can see is clearly a straight line and if t = 0, [A] = C, which implies C is the initial concentration of A.
If we consider what happens if the reaction is now first order with respect to A (x = 1), we have
1/[A] d[A] = -k dt (since anything to the power of -1 is its reciprocal).
Again, integrating this:
ln[A] = C - kt
If we now take the exponential function of this, we have
[A] = e^(C - kt) = e^C x e^-kt
Which is clearly an exponential decay. Again, if t = 0, then [A] = e^C, so e^C is the initial concentration of A.
If we now consider the case where the reaction is second order wrt A (e.g. x = 2), you get the idea what’s going to happen next - we substitute, integrate, rearrange as appropriate and then interpret how the constant of integration C relates to the initial concentration of A.
1/[A]^2 d[A] = -k dt => -1/[A] = C - kt
[A] = 1/(kt - C) = 1/(kt + c) (where c = -C)
When t = 0, [A]_0 = 1/c, so c = 1/[A]_0, where [A]_0 is the initial concentration of A.
Of course, you are NOT expected to reproduce any of the above for A level chemistry. Nor do you need the results - this just shows why a zero order reaction is a straight line and the others are curves.
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