I just wanted to check my understanding of this. Given the reaction,

$$ (1) \; \ce{A->B} $$

the corresponding rate law is $ v=k[A] $

Also consider

$$ (2) \; \ce{A <=> B} $$

with forward rate law: $ v_1=k_1[A] $ and reverse: $ v_2=k_2[B] $. This makes the corresponding rate equation $ v=v_1-v_2=k_1[A]-k_2[B] $.

My question is in the difference between the two. I've often seen reactions expressed in both ways. In $ (1) $, is the rate law assuming the reverse reaction not to occur (or at least to be completely negligible? That's why I drew it with $ \ce{->} $ instead of $ \ce{<=>} $. Am I correct in my thinking? In actuality, is it always more correct to use the rate law of $ (2) $? The rate law for $ (1) $ could also correspond to initial reaction velocity, right? This is because there is no $ B $ present at the start of the reaction.

If this is correct, then the integrated rate equations will become substantially more complicated using rate laws of form $ (2) $. I can't use simple equations like $ [A]_t=[A]_0 e^{-kt} $, as this wouldn't take into account the reverse reaction.


Not all reactions can or should be represented with a double arrow. The general-audience example is that you "can't unbake a cake". You are right that there are instances that we assume that k2 is negligible (relative to k1) and therefore a reaction is treated as if it is reversible.

Here's a sample plot of how the concentrations of A for the irreversible and reversible reactions compare for k1 = k2 = 1.

Mathematica graphics

No surprises here; the reversible reaction results in an equilibrium concentration of A. We can compare the concentration of A in each of the reactions as a function of kr:

Mathematica graphics

Here I've plotted Log[k2] on the x-axis and the ratio of A in the irreversible reaction to the reversible reaction. The different lines are at various times (in seconds). I've kept k1 = 1 and the initial concentration 1 as well. The two equations predict the same concentration of A in the limit of very small k2, and by very small I suspect this means at least a factor of 10^4 or more (not shown on the graph).

While you are correct that the integrated rate laws get much more complicated for reversible reactions, the one you are considering is still solvable in closed form.


In $(1)$ the reaction is irreversible. For example, potassium chlorate decomposes on heating to form potassium chloride and oxygen:
$$\ce{2KClO3 -> 2KCl + 3O2}$$ In this reaction, the products cannot combine to give back potassium chlorate (I think the reason would be because of reaction not being performed in closed container). In case of irreversible reactions, the reactions occur only in forward direction. In actuality, you can use rate equation of $(2)$ for the above reaction, as rate backward reaction is zero, you will end up with rate equation of $(1)$.

Velocity of reaction $(1)$ doesn't corresponds to velocity of reaction at time $t=0$. It is the general equation for velocity of the reaction.

Consider a first order reaction $$\ce{A -> products}$$
Velocity of the reaction is given by $\frac{\mathrm dx}{\mathrm dt}=k[\ce{A}]$. Here, $[\ce{A}]$ corresponds to concentration of reactant $\ce{A}$ at a particular time and it doesn't correspond to concentration of $\ce{A}$ at $t=0$.

To understand it consider the following explanation. let the initial concentration of the reactant be $a\ \mathrm{mol\ dm^{-3}}$. Let $x\ \mathrm{mol\ dm^{-3}}$ decompose in $t$ seconds. The remaining concentration of the reactant is $a-x\ \mathrm{mol\ dm^{-3}}$.

${\therefore}$ Velocity of the reaction after a time 't' is given by $\frac{\mathrm dx}{\mathrm dt}=k[a-x]$.

Rate equations for reversible reactions will be different as given in this page (see the first order reaction part).


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