3
$\begingroup$

$$\ce{2A -> 3B + C + 2D}$$

This is a first order reaction. Rate constant $k$ for reaction is given to be (say) $\pu{1.386E-2 min^{-1}}$. From this we get the value of $t_{\frac{1}{2}}$ of reaction as $\frac{\ln(2)}{k} = \pu{50 min}$. Now this $t_{\frac{1}{2}}$ should mean half life time of reaction (and not for $\ce{A}$). However, in my textbook it is written that if concentration of $\ce{A}$ is $\pu{200 M}$ at $t=0$ then at $t = \pu{100 min}$ the conc. of $A$ would be $\pu{50 M}$ as $100$ minutes is two times of the half life (and thus the conc. of $A$ should reduce by $2^2$ times).

I think this is wrong as the half life of $A$ is half of the half-life of reaction (since rate of consumption of $A$ is double the rate of reaction as can be seen from the stoichiometric coefficients). According to me after $100$ minutes the concentration of $A$ should be $\pu{25 M}$. Is my logic correct? Or am I missing something?

$\endgroup$
3
$\begingroup$

Consider what happens to your argument if I rewrite the equation in the following equivalent form: $$\ce{A -> \frac{3}{2}B +\frac{1}{2}C +D}$$ That is, you write

... since rate of consumption of A is double the rate of reaction as can be seen from the stoichiometric coefficients

In contrast, the rate of the reaction, in this case, is defined to be the rate of decay of $\ce{A}$, $$-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k[\ce{A}]$$ When you look at it like this, it is more clear that the stoichiometry should not have an effect on the decay process as you claimed.

Furthermore, if we integrate the equation above after separating by variables, $$\int_{[\ce{A}_0]}^{[\ce{A_t}]}\frac{1}{[\ce{A}]}\mathrm{d}[\ce{A}] = \int_0^t-k\mathrm{d}t$$ we get, $$\ln\left(\frac{[\ce{A_t}]}{[\ce{A}_0]}\right)=-kt.$$

If we try to find the time at which the original concentration has halved, we get the formula you have written above which is, $$t_{\frac12}=\frac{\ln(2)}{k}.$$

The important point here, and this isn't true for other reaction-orders, is that the half-life is independent of initial concentration. This is not what one would intuitively expect, and indeed it's not true in general, but it is characteristic of all exponential-decay processes.

Indeed, this is a very deep result which doesn't tend to get emphasized enough in my opinion.

It is true that sometimes you can say that a reaction will be second order based on stoichiometric coefficients, but you are told this is first order, so that is probably the source of your confusion.

So, your textbook is right.

$\endgroup$
  • $\begingroup$ You cannot change the stoichiometric coefficient of the reaction. "but you are told this is first order"....order has nothing to do with stoichiometric coefficient.. The rate constant $k$ had been given for the first equation i.e. $2A \rightarrow 3B+C+2D$ and not for $A \rightarrow (3/2)B+(1/2)C+D$. $\endgroup$ – user38977 Apr 18 '17 at 4:17
  • 3
    $\begingroup$ Stoichiometric coefficients only tell you the proportion of reactant molecules to product molecules. For all chemical reactions, you can divide through by a constant and it changes absolutely nothing about the actual reaction taking place. If it happens that the reaction is two molecules of $A$ colliding, then it makes less sense physically to write it that way, but all rate equations, etc. derived from the reaction will be identical. In contrast, because this is a first order process, it makes more physical sense to write the equation down with a coefficient of $1$ in front of $A$. $\endgroup$ – jheindel Apr 18 '17 at 5:16
  • 1
    $\begingroup$ The reason I say the whole thing about being told it is first order is because it is often tempting to guess that because there is a $2$ in front of the $A$, it is a bimolecular reaction, but it need not be one as we see here. $\endgroup$ – jheindel Apr 18 '17 at 5:17
  • 1
    $\begingroup$ @blue, it may be useful to distinguish elementary steps from balanced reactions. Elementary steps are those that directly represent molecular interactions, and those for which coefficients cannot be so manipulated. Balanced reactions, however, simply provide a summary of the reaction, and make no attempt to specify the mechanism or interactions taking place. $\endgroup$ – a-cyclohexane-molecule Apr 18 '17 at 5:22
  • 3
    $\begingroup$ @blue, I think I see your point. Your issue isn't with reaction order, but the rate constant definition. To summarize your argument: say our reaction is $\ce{A -> B}$ and $k$ = 10 s$^{-1}$, but if our reaction is $\ce{2A -> 2B}$, then $k$ would be 5 s$^{-1}$. The half-life of species A remains the same in either case, being $t_{1/2} = \ln 2/k$ in the former and $t_{1/2} = \ln 2/2k$ in the latter, but the reaction rate constant is different. Is that right? I think your argument is sound, but I'd like to see what others have to say. $\endgroup$ – a-cyclohexane-molecule Apr 18 '17 at 6:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy