# Time dependence of integrated rate equations

Is it possible for an integrated rate equation to not be dependent on time? What law or principle says so in chemical kinetics?

What if an integrated rate equation obtained only depends on the concentrations of the reactants? Is it possible? Can you refer me to books and references?

By definition, an integrated rate equation shows the change in concentration as a function of the progress of the reaction (time). So I don't understand why professor insists that an integrated rate equation can exist without having the variable time in it.

• If that's really what the Prof. said then it doesn't make any sense. The only special case which is usually covered is that the half-life of a first order reaction is not dependend on concentration. But that's a totally different statement so I don't think that's what was said/he wanted to say.
– DSVA
Dec 6 '16 at 9:48
• Hello, if you look at the derivation posted by the thread starter at chemistry.stackexchange.com/questions/62495/… , you will see that at the end, he obtained an equation independent of time and so, he had an integrated rate equation which is not a function of time. However, I think that this is fundamentally wrong but I don't have a strong argument or a reference that explicitly tells that this is wrong. That's why I cannot defend my stand. Dec 6 '16 at 9:59
• Under steady state conditions it is assumed that $d[C]/dt =0$ so time is removed this way. However, this only applies under certain conditions and only for some reactions. Secondly, if an equilibrium is involved, at long times when equilibrium is established concentrations are effectively constant. Other than these cases the concentration of all species vary with time. Dec 6 '16 at 16:43
• Hey porphyin. Yeah by that condition the concentration of C does not depend on time, but the reactants A and B still do. I think you agree that the concentration of reactants are still time dependent and thus, an integrated rate equation without time variable (except when already in equilibrium) is problematic Dec 6 '16 at 22:04

From a mathematical standpoint: If the integrated rate law is independent of time, then the rate of change of the concentration must equal zero (by the fundamental theorem of calculus). So you're looking for a reaction in which the concentration of the species you want in the rate law (presumably the reactant) does not change appreciably.

As others pointed out, this is a common case for intermediates under the steady-state approximation (rate of production = rate of consumption, net rate zero). But that's just for intermediates. For a substrate, you would need the substrate concentration to be essentially constant. That's probably not a "reaction" that will be interesting to study. A trivial solution, if you will.

Ok, so if we got the reaction

$A\rightleftharpoons B$

with $k_1$ for $A \rightarrow B$ and with $k_2$ for $B \rightarrow A$ then

$\frac{d[A]}{dt}=-k_1 [A]+ k_2 [B]$

if you integrate that you end up with

$k_1 [A]=k_2 [B]$

which is time independent

• This is making the assumption that the initial concentration of [A] is the equilibrium concentration. If the system is initially not at equilibrium then [A] varies with time: i.stack.imgur.com/L7X6c.jpg Dec 6 '16 at 10:26
• yes ofc. I wanted to write that that's the case after the equilibrium is reached, somehow forgot that.
– DSVA
Dec 6 '16 at 17:01
• So is equilibrium only the case wherein reaction rate is time independent? Since effectively the change in concentration of all reacting species per unit time is zero, right? Dec 6 '16 at 22:06
• @DSVA I haave a trivial question about integrated rate laws...so trivial that I searched for "integrated rate laws" and found this question relevant, so to comment. why is the rate law integrated than to differentiated? R is the instantaneous rate of concentration function against Time. R is a derivative. why is it integrated then? what is the intention? Jul 23 '17 at 1:46