Question

1. For the reaction 3A — C, the initial concentration of A was 0.2 M, and the reaction rate was
1.0 M/s. When [A] was doubled, the reaction rate increased to 4.0 M/s. Determine the rate
law for the reaction.

Answer 1

Answer:

$r=25M^{-1}s^{-1}[A]^2$

Explanation:

Hello there!

In this case, according to the given information for this chemical reaction, it is possible for us to set up the following general rate law and the ratio of the initial and the final (doubled concentration) condition:

$r=k[A]^n\\\\\frac{r_1}{r_2} =\frac{k[A]_1^n}{k[A]_2^n}$

Next, we plug in the given concentrations of A, 0.2M and 0.4 M, the rates, 1.0 M/s and 4.0 M/s and cancel out the rate constants as they are the same, in order to obtain the following:

$\frac{1.0}{4.0} =\frac{0.2^n}{0.4^n}\\\\0.25=0.5^n\\\\n=\frac{ln(0.25)}{ln(0.5)} \\\\n=2$

Which means this reaction is second-order with respect to A. Finally, we calculate the rate constant by using n, [A] and r, to obtain:

$k=\frac{r}{[A]^n} =\frac{1.0M/s}{(0.2M)^2}\\\\k=25M^{-1}s^{-1}$

Thus, the rate law turns out to be:

$r=25M^{-1}s^{-1}[A]^2$

Regards!