3
$\begingroup$

Consider a parallel reaction

$\ce{A->B}$ (rate constant for this reaction is $k_1$ and it follows first order kinectics),

$\ce{2A->C}$ (rate constant for this reaction is $k_2$ and it follows first order kinetics).

Then we can write $$\mathrm dA/\mathrm dt = \mathrm dA_1/\mathrm dt + \mathrm dA_2/\mathrm dt$$ (where $\mathrm dA_1$ is the amount of $\ce A$ going through the first path and $\mathrm dA_2$ is the amoubt of $\ce A$ going through 2nd path),

$$\mathrm dA/\mathrm dt = k_1[\ce A] + 2k_2[\ce A]$$

Then to calculate the $k$ where $k$ is the rate constant for the overall reaction we write

$$k[\ce A] = k_1[\ce A] + 2k_2[\ce A]\tag i$$

And so $k$ comes out to be $k_1+2k_2$.

My doubt is that to write $-\mathrm d[\ce A]/\mathrm dt = k[\ce A]$ did we assume that in the overall reaction the coefficient of $\ce A$ in the overall reaction to be 1.

Taking a general case $$\ce{$a$A->products}$$
we can write $$-\mathrm d[\ce A]/\mathrm dt=ak[\ce A]$$

But according to the equation $\text{(i)}$ we took $a=1$. Why?

If suppose $\ce A$ is a radioactive substance and we have to calculate the mean life of $\ce A$ then we require to know the overall rate constant, so foe that what should be the value of $a$?

$\endgroup$
  • $\begingroup$ Your question is poorly defined. Are both reactions $\ce{A -> B}$ and $\ce{2A -> C}$ first-order with respect to $\ce{A}$, or are they elementary steps (which means you are missing a power of two)? The second equation is missing a minus sign since the LHS is negative and the RHS is positive. Lastly, the stoichiometric coefficient of $\ce{A}$ does not affect the order of the reaction wrt $\ce{A}$. If your "overall reaction" (which btw isn't even defined) has a rate law of $k_1[\ce{A}] + k_2[\ce{A}]^2$ then that is just how it is. You cannot simply label it as "first-order" or "second-order". $\endgroup$ – orthocresol Oct 5 '15 at 18:35
  • $\begingroup$ Both the parallel reactions are of first order type $\endgroup$ – user2215860 Oct 6 '15 at 2:45
  • $\begingroup$ In the second equation the lhs is negative and so is the rhs as d[A]1/dt and d[A]2/dt are also negative as they are also decreasing.It is not the order which I am confused about,it is the overall rate constant which I am concerned as if A is a radioactive substance and we have to calculate its mean life we require the overall rate constant. $\endgroup$ – user2215860 Oct 6 '15 at 2:53
3
$\begingroup$

The key to answering these kinds of riddles is to think about units. Rates of reaction have units of $\mathrm{\frac{mol}{L\cdot s}}$, but that' isn't fully specified. You need to know moles of what.

In your case, the confusion is stemming from confusion of what the "moles" in $\mathrm{\frac{mol}{L\cdot s}}$ refers to.

The answer can either be moles of "reaction", or moles of A. The other key thing to realize is that you can write "the" reaction however you want. The value of the "rate constant" $k$ depends on how you define the rate. If for the overall reaction you write:

$$\ce{3A -> \varnothing}$$

...and if the "rate" you are calculating is the rate of reaction, then there are three moles of A lost per mole of reaction, so

$${dA \over dt}=-3k_{\alpha}\ce{[A]}$$

If instead you decide "the" reaction is

$$\ce{A -> \varnothing}$$

then you can rate the rate of disappearance of A as

$${dA \over dt}=-k_{\beta}\ce{[A]}$$

The choice is up to you and it is arbitrary. You just have to keep track of the moles of what the rate laws you write refer to.

Then to calculate the k where k is the rate constant for the overall reaction we write [[Equation (i)]]

You need to think carefully about what you meant by the words "the overall reaction". When you write equation 1, you are adding together rates of loss of A, i.e. the units are $\mathrm{\frac{mol~of~A}{L\cdot s}}$. The only reaction you can write that has the same rate is one where the coefficient of A is 1.

"The" half-life for radioactive elements usually refers to units where units are $\mathrm{\frac{mol~of~element}{s}}$. The fact that the unit is $\mathrm{\frac{mol~of~element}{s}}$ and not $\mathrm{\frac{mol~of~reaction}{s}}$ means that if we wanted to write the corresponding chemical reaction, the stoichiometric coefficient had better be 1.$%edit$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.