# For the reaction 2"NO"(g) + "Br"_2(g) -> 2"NOBr"(g), the following data were obtained. What is the rate constant?

## ul("Trial"" "["NO"]" "" "["Br"_2]" "color(white)("///")r(t) ("M/s")) 1color(white)("//////")"0.10 M"color(white)("///")"0.20 M"color(white)(///)24 2color(white)("//////")"0.25 M"color(white)("///")"0.20 M"color(white)(///)150 3color(white)("//////")"0.10 M"color(white)("///")"0.50 M"color(white)(///)60

Mar 26, 2017

The general rate law would be

$r \left(t\right) = k {\left[A\right]}^{m} {\left[B\right]}^{n}$

$= - \frac{1}{a} \frac{\Delta \left[A\right]}{\Delta t} = - \frac{1}{b} \frac{\Delta \left[B\right]}{\Delta t} = \frac{1}{c} \frac{\Delta \left[C\right]}{\Delta t} = \frac{1}{d} \frac{\Delta \left[D\right]}{\Delta t}$,

for the reaction

$a A + b B \to c C + \mathrm{dD}$,

where:

• $r \left(t\right)$ is the initial rate of reaction.
• $k$ is the rate constant for a particular reaction at a particular temperature.
• $\left[X\right]$ is the initial concentration of $X$.
• $m$ and $n$ are orders with respect to $A$ and $B$, respectively. That is, they are the exponents that describe the contribution of $A$ or $B$ to the overall rate.

Therefore, the rate law for the reaction

$2 N O \left(g\right) + B {r}_{2} \left(g\right) \to 2 N O B r \left(g\right)$

is:

$r \left(t\right) = k {\left[N O\right]}^{m} {\left[B {r}_{2}\right]}^{n}$

An initial rate ${r}_{i} \left(t\right)$ can be written for any trial $i$. Thus, one can write the initial rate for trials 1, 2, 3, . . . n, as convenient:

${r}_{1} \left(t\right) = k {\left[N O\right]}_{1}^{m} {\left[B {r}_{2}\right]}_{1}^{n}$

${r}_{2} \left(t\right) = k {\left[N O\right]}_{2}^{m} {\left[B {r}_{2}\right]}_{2}^{n}$

${r}_{3} \left(t\right) = k {\left[N O\right]}_{3}^{m} {\left[B {r}_{2}\right]}_{3}^{n}$

Since we do not know the rate constant, we can simply ignore it for the moment by dividing the rates. It is convenient to choose trials such that one of the concentrations remains constant across trials, eliminating a variable.

$\frac{{r}_{1} \left(t\right)}{{r}_{2} \left(t\right)} = \frac{\cancel{k} {\left[N O\right]}_{1}^{m} {\left[B {r}_{2}\right]}_{1}^{n}}{\cancel{k} {\left[N O\right]}_{2}^{m} {\left[B {r}_{2}\right]}_{2}^{n}}$

$= {\left(\frac{{\left[N O\right]}_{1}}{{\left[N O\right]}_{2}}\right)}^{m} {\cancel{{\left(\frac{{\left[B {r}_{2}\right]}_{1}}{{\left[B {r}_{2}\right]}_{2}}\right)}^{n}}}^{1}$

We eliminated the bromine concentrations since ${1}^{n} = 1$. Thus, we can solve for the order $m$ with respect to $N O \left(g\right)$. If the ratios are not nice, then:

$\ln \left[\frac{{r}_{1} \left(t\right)}{{r}_{2} \left(t\right)}\right] = \ln {\left(\frac{{\left[N O\right]}_{1}}{{\left[N O\right]}_{2}}\right)}^{m}$

$= m \ln \left(\frac{{\left[N O\right]}_{1}}{{\left[N O\right]}_{2}}\right)$

$\implies m = \frac{\ln \left[\frac{{r}_{1} \left(t\right)}{{r}_{2} \left(t\right)}\right]}{\ln \left(\frac{{\left[N O\right]}_{1}}{{\left[N O\right]}_{2}}\right)}$

However, your ratios are nice. From trials 1 and 2, $\frac{{\left[N O\right]}_{1}}{{\left[N O\right]}_{2}} = 0.4$, while $\frac{{r}_{1} \left(t\right)}{{r}_{2} \left(t\right)} = 0.16$. We can instead say,

"What is $0.4$ raised to so that it equals $0.16$?" That is, "what exponent makes it so that the ratio of the concentrations equals the ratio of the rates?"

Naturally, the answer is $2$, i.e. ${0.4}^{2} = 0.16$. One can also get the same answer by brute force logarithm calculations:

$\textcolor{g r e e n}{m} = \frac{\ln 0.16}{\ln 0.4} = \ln \frac{{0.4}^{2}}{\ln \left(0.4\right)} = \frac{2 \cancel{\ln 0.4}}{\cancel{\ln 0.4}} = \textcolor{g r e e n}{2}$

For the other order $n$ with respect to $B {r}_{2}$, simply choose trials 1 and 3, since those hold $\left[N O\right]$ constant. We analogously have that ${0.4}^{n} = 0.4$, and thus $n = 1$.

Or the brute force way:

$\textcolor{g r e e n}{n} = \frac{\ln \left[\frac{{r}_{1} \left(t\right)}{{r}_{3} \left(t\right)}\right]}{\ln \left(\frac{{\left[B {r}_{2}\right]}_{1}}{{\left[B {r}_{2}\right]}_{3}}\right)} = \frac{\ln 0.4}{\ln 0.4} = \textcolor{g r e e n}{1}$

Therefore, the rate law becomes:

$\textcolor{g r e e n}{r \left(t\right) = k {\left[N O\right]}^{2} \left[B {r}_{2}\right]}$

Finally, we have the info to plug and chug. Choose any trial you wish. You will get effectively the same $k$. I choose $1$.

$\textcolor{b l u e}{k} = \frac{{r}_{1} \left(t\right)}{{\left[N O\right]}_{1}^{2} {\left[B {r}_{2}\right]}_{1}}$

= ("24 M/s")/(("0.10 M")^2("0.20 M"))

$= \textcolor{b l u e}{1.2 \times {10}^{4}}$

What must the units be?