# Determine the Value and Units of the Rate Constant

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

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

[NOBr]o(M)    Initial rate of formation of Br2 (M/s)
0.600         1.08 * 10^2
0.750         1.69 * 10^2
0.900         2.43 * 10^2


I'm not quite understanding 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". – M.A.R. Jan 31 '15 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. – JeannaT Jan 31 '15 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. – M.A.R. Jan 31 '15 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. – Ben Norris Jan 31 '15 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]$^2$