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.

  • 2
    $\begingroup$ 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". $\endgroup$
    – M.A.R.
    Jan 31, 2015 at 19:58
  • $\begingroup$ 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. $\endgroup$
    – JeannaT
    Jan 31, 2015 at 20:07
  • $\begingroup$ 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. $\endgroup$
    – M.A.R.
    Jan 31, 2015 at 20:11
  • $\begingroup$ 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. $\endgroup$
    – Ben Norris
    Jan 31, 2015 at 20:27

2 Answers 2


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.


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


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