Question

What is the value of the rate constant k for this reaction?When entering compound units, indicate multiplication of units explicitly using a multiplication dot (multiplication dot in the menu). For example, M−1⋅s−1.Express your answer to two significant figures and include the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot or a dash.

Answer 1

The table with the data is in the picture attached.

Answer:

• $k=0.0033M^{-2}.s^{-1}$

Explanation:

The reaction equation suggests that the law could have this form:

• $rate=k[A]^a[B]^b[C]^c$

Then, the work is to find the values of the exponents that satisfy the initial rate data.

A first glance shows that for the third and fourth trials the initial rates are the same. Since for these two trials only the initial concentration of substance B changed (A and C were kept equal), you conclude that the reaction rate does not depend on B, and ist exponent (lower b) is 0.

Then, so far you can say:

• $rate=k[A]^a[C]^c$

When you use trials 1 and 2, you get:

$\frac{r_2}{r_1}=\frac{27M/s}{9M/s}=\frac{(0.3M)^a(0.3M)^b(0.9M)^c}{(0.3M)^a(0.3M)^b(0.3M)^c}=3^c\\\\ 3=3^c\\ \\ 1=c$

Now, you can use trials 1 and 3 to determine the other exponent:

$\frac{r_3}{r_1}=\frac{36M/s}{9M/s}=\frac{(0.6M)^a(0.3M)^b(0.3M)^c}{(0.3M)^a(0.3M)^b(0.3M)^c}=2^a\\\\ 4=2^a\\ \\ 2^2=2^a\\ \\ 2=a$

Thus, you have the rate law:

• $r=k[A]^2[C]$

Now, you just use any trial to obtain k. Using trail 1:

• $k(0.3M)^2(0.3M)=9.10^{-5}M/s$

Which yields:

• $k=0.0033M^{-2}.s^{-1}$