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  1. The reactant concentration in a zero-order reaction was $7.00×10^{−2}~\mathrm{M}$ after $160~\mathrm{s}$ and $2.50×10^{−2}~\mathrm{M}$ after $345~\mathrm{s}$. What is the rate constant for this reaction?
  2. What was the initial reactant concentration for the reaction described in part one?
  3. The reactant concentration in a first-order reaction was $7.70×10^{−2}~\mathrm{M}$ after $50.0~\mathrm{s}$ and $1.60×10^{−3}~\mathrm{M}$ after $85.0~\mathrm{s}$. What is the rate constant for this reaction?

Background info: The integrated rate laws for zero and first order reactions may be arranged such that they resemble the equation for a straight line, $y=mx+b$.

\begin{array}{lccc} \text{Order}& \text{Integrated Rate Law}& \text{Graph}& \text{Slope} \\\hline 0 & [A]=−kt+[A]_0 & [A]\text{ vs. }t & −k\\ 1 & \ln[A]=−kt+\ln[A]_0 & \ln[A]\text{ vs. }t & −k\\\hline \end{array}

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    $\begingroup$ Hint - For (a) how many points does it take to define a straight line? That should get you started. $\endgroup$ – MaxW Nov 6 '15 at 1:44
  • $\begingroup$ I'm still lost :( $\endgroup$ – muhu Nov 6 '15 at 1:53
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    $\begingroup$ Draw X/Y axis. Label X as seconds and Y as concentration. For (a) calculate the slope of line. $\endgroup$ – MaxW Nov 6 '15 at 2:03
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According to your table, for a zero order reaction, you make a graph of the measured concentrations vs the corresponding times. It will be a straight line, and the slope and intercept will give you the rate constant and the concentration at time zero.

For a first order reaction, you make a graph of the natural log of the concentrations vs the corresponding times. It will be a straight line, and the slope and intercept will give you the rate constant and the concentration at time zero.

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Question 1

First, note the integrated rate law for zero-order reaction: $$[A] = -kt + [A]_\circ$$ As you have already realised, this is in the form of a linear equations: $$y = mx + b$$ where $y$ corresponds to [A],

$m$ which is the gradient correpsonds to $-k$

$b$ which is y-intercept corresponds to $[A]_\circ$

The question is asking for the rate constant, so all we have to do is find the gradient of this line and that will be equal to $-k$. $$m = \frac{[A]_2 - [A]_1}{t_2 - t_1} = \frac{7.00\times 10^{-2}\ \mathrm{M} - 2.50\times 10^{-2}\ \mathrm{M}}{160\ \mathrm{s} - 345\ \mathrm{s}} = -2.43\times 10^{-4}\ \mathrm{M\ s^{-1}}$$ Therefore $k = 2.43\times 10^{-4}\ \mathrm{M\ s^{-1}}$

Question 2

Now that we know what $k$ is, we have a equation where there is 4 variables and we know 3 of them. Therefore, all we have to do is substitute the values that we know and then solve for the unknown variable, $[A]_\circ$. $$[A] = -kt + [A]_\circ$$ $$[A]_\circ = [A] + kt$$ $$[A]_\circ = 2.50\times 10^{-2}\ \mathrm{M} + 2.43\times 10^{-4}\ \mathrm{M\ s^{-1}}\times 345\ \mathrm{s} = 1.09\times 10^{-1}\ \mathrm{M}$$ Therefore the initial concentration of the reactant is: $1.09\times 10^{-1}\ \mathrm{M}$

Question 3

This question is exactly the same as question 1, except the only difference is that the reaction is first order, not zero order. To solve this question, just follow the same procedure that we used to do question 1. The only difference is going be is that the integrated rate law is going to be slightly different.

I will leave this question for you do. If you have any problems feel free to ask.

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