For a chemical reaction

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

the order of reaction is one with respect to $\ce{A}$ and one with respect to $\ce{B}$. The initial rate of the reaction is $\pu{4e-2 mol L-1 s-1}.$ When $50\%$ of the reactants are converted into products, the rate of the reaction would become

(A) $\pu{2e-2 mol L-1 s-1}$
(B) $\pu{1e-2 mol L-1 s-1}$
(C) $\pu{4e-2 mol L-1 s-1}$
(D) $\pu{2e-1 mol L-1 s-1}$

This is the first time I am attempting kinetics problems. Is the question asking the concentration when the system has reached its half-life?

Could I use the half-life formula here?

My attempt at a solution:

I used the rate law which I equated to the definition of rate.

I have serious conceptual issues which I hope to clear up. So is this this right?

$$\frac{1}{c(t)} - \frac{1}{c_0} = kt$$

Would this formula solve the above problem?


Is this the right formula to use to solve the above problem

Yes, this is it, if the initial concentrations of the reagents are the same (which is not mentioned explicitly in the problem statement, but probably is implied, because otherwise the problem would be unsolvable.)

Indeed, as A and B are spent in equal amounts, their concentrations are bound to be the same not only initially, but at all times: $[\ce{A}]=[\ce{B}]=c(t)$. Therefore $\dot c=-k\cdot[\ce{A}]\cdot[\ce{B}]=-k\cdot c^2$, which you seem to already know how to solve.

Update. Come to think of it, you don't actually need to solve for $c(t)$. You know that the rate is proportional to the first power of $[\ce{A}]$, and also to the first power of $[\ce{B}]$, and that both $[\ce{A}]$ and $[\ce{B}]$ have decreased by 50%; what would happen to the rate?

The concept of half life does not apply here, since the decay is not exponential.

Also, welcome to Chem.SE, and pay attention to the formatting.

  • 1
    $\begingroup$ Is the answer $ 1 * 10^{-2}$ $\endgroup$
    – SID
    May 22 '16 at 10:50
  • $\begingroup$ Looks like so. The concentrations decrease by 50%, and the rate depends on $c^2$, so it must have become 4 times lower. $\endgroup$ May 22 '16 at 10:53

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