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A+B ———> C+D

If A and B are taken in excess then what will be the order of reaction?

I thought since it is taken in excess, their change wont really be significant and so answer would be 0.

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    $\begingroup$ Having both A and B in excess makes no sense. There are only three possible cases. (1) A > B, (2) A = B, or (3) A<B. If (1) Then A is in excess. If (3) then B is in excess. $\endgroup$
    – MaxW
    Nov 11, 2018 at 18:55
  • $\begingroup$ Reaction orders are not related to excesses at all. The question is wrong in its premises. $\endgroup$ Nov 11, 2018 at 19:23
  • $\begingroup$ No like pseudo kind of rate order@IvanNeretin $\endgroup$
    – Sweetner
    Nov 11, 2018 at 19:25
  • $\begingroup$ If the reaction is a kind of heterogeneous catalytic reaction or photochemical reaction your answer correct $\endgroup$ Nov 15, 2018 at 7:22

2 Answers 2

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3 doesn't make any sense for the reaction order.

Given the elementary reaction

$$\ce{aA + bB -> \text{products}}$$

then the rate equation would be

$$r = k\ce{[A]^a[B]^b}$$

since by definition an elementary reaction is a reaction for which the stoichometric coefficients are the same as exponents for the concentrations in the rate equation. Thus the order of the reaction is $a+b$.

So for the case of the elementary reaction

$$\ce{A + B -> \text{products}}$$

$a=1$ and $b=1$ so the reaction order is 2.

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Reaction orders are defined per component. If both A and B are in excess (but available in roughly the same concentration), the reaction order would be one in either of them.

If A or B is in excess, then the reaction order is one for the compound in excess and zero for the other compound.

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