## Help with these kinetics questions about reaction order with respect to one reactant A??

Answer:

Highlight to reveal answers once you’ve figured it out.

1) Well, consider the following half-lives:

In all of these, only the zero order half-life is directly proportional to the concentration of \(A.\) Hence, the reaction is zero order with respect to \(A.\)

2) You can figure out a lot from the rate law…

\(r(t)=k[A]^{n}\).

Knowing that doubling the concentration leads to quadrupling the rate, i.e. \(2[A] \rightarrow 4 r(t)\), we have that

\(4 \cdot r(t)=k(2[A])^{n}=2^{n} \cdot k[A]^{n}\).

Thus, we have that \(2^{n}=4 \text {. What must } \mathbf{n} \text { be? }\) The reaction order with respect to \(A\) is of \(this\) nth order.

3) Same process as \((2)\). Now we claim that:

\(1.41 \cdot r(t)=k(2[A])^{n}=2^{n} \cdot k[A]^{n}\).

But \(1.41 \approx \sqrt{2}\). Hence, we have that \(\sqrt{2}=2^{n} \text {. What is } \mathbf{n} \text { this time? }\) (What is \(2\) raised to that correlates with a square root?) The reaction is of this nth order with respect to \(A.\)

4) It discusses the first and second half-lives and claims that they are equal. This just says that the half-life (of THIS order) does not depend on what concentration you start at.

Look above. Which order half-life does not depend on \([A]_{0} ?\) Hint: it’s not a prime number.

5) This says that the rate of reaction does not depend on the current \([A]\). That is, \(A\) has no influence on the rate. That can only be the case if \(A\) is zero order, i.e. the rate depends only on the rate constant:

\(r(t)=k[A]^{0}=k\)