# Determining the value and units of the rate constant

Consider this initial-rate data at a certain temperature for the reaction described by

$$\ce{2NOBr(g) -> 2NO(g) + Br2(g)}$$

$$\begin{array}{cc}\hline \ce{[NOBr]_0 (M)} & \mathrm{Initial \,rate \, of \,Br_2}\, (\pu{M/s}) \\ \hline 0.600 &1.08 \times 10^2 \\ 0.750 & 1.69 \times 10^2\\ 0.900 & 2.53 \times 10^2\\ \hline \end{array}$$

I'm not understanding how to find initial rates from the data given and then use that to find the rate constants. I also don't really understand how this information can give me the orders of reactions, for example when does the number double, stay the same, or quadruple, etc.

• Welcome to chemistry.SE! This is a homework question. Thus, we should make sure that we aren't doing homework for you. You should provide some info so that we make sure you're "aware of the underlying concepts". Jan 31, 2015 at 19:58
• I guess I'm not quite understanding to find initial rates and rate constants. I also don't really understand the orders of reactions and which number and when does the number double, stay the same, or quadruple. Jan 31, 2015 at 20:07
• Hmm. I think I'll leave this for the community to decide. Maybe a good answerer will be able to do this with hints. I'll also edit your question to include a "reference-request", so that the answer will include a page link for more studying. Jan 31, 2015 at 20:11
• I've moved key ideas from your comment, @JeannaT, into your question, which gives folks a place to start on this question without it needing closed. Jan 31, 2015 at 20:27

Alright, so in the reaction

$$\ce{2NOBr(g) -> 2NO(g) + Br2}$$

the rate law is

$$\text{rate} = k[\ce{NOBr}]^x$$

where $k$ is some constant and $x$ is the order of the reaction in respect to $\ce{NOBr}$.

By seeing how the initial rate changes when we change the concentration of $\ce{NOBr}$, we can determine the value of $x$. We can use any two of the three. I'm going to use the first and third trials. If we divide them we get

$$\frac{\text{rate 3}}{\text{rate 1}} = \frac{k[\ce{NOBr}]_3^x}{k[\ce{NOBr}]_1^x}$$

$$\frac{2.43\times10^2}{1.08\times10^2} = \frac{0.900^x}{0.600^x}$$

The $k$'s cancel out.

$$2.25 = 1.5^x$$

$$x = 2$$

The rate is second order in respect to $\ce{NOBr}$, and the rate law is written $\text{rate} = k[\ce{NOBr}]^2$. If you double the concentration, the rate will quadruple.

$$\text{rate before doubling concentration} = k[\ce{NOBr}]^2$$

\begin{align}\text{rate after concentration} &= \left(2[\ce{NOBr}]\right)^2\\ &= 2^2[\ce{NOBr}]^2\\ &= 4[\ce{NOBr}]^2\\ &= 4 \times \text{rate before doubling concentration}\end{align}

A tripling of the concentration will increase the rate by a factor of nine, a quadrupling of the concentration increases the rate by a factor of 16, and so on.

The ratio of 0.9 to 0.6 is 1.5.

The ratio of 2.43 to 1.08 is 2.25.

$1.5^2=2.25$

The initial rate is proportional to [NOBr]$^2$