## Q 5A 2014 FINAL

Palmquist_Sierra_2N
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### Q 5A 2014 FINAL

Question 5A on the winter 2014 final in the back of our course reader solves for the rate constant at the end but doesn't show how. I understand how to find which order the reaction is and the rate law but not how to find the rate constant with the given information. Please help/explain. Thanks!

ntruong2H
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### Re: Q 5A 2014 FINAL

Palmquist_Sierra_2N wrote:Question 5A on the winter 2014 final in the back of our course reader solves for the rate constant at the end but doesn't show how. I understand how to find which order the reaction is and the rate law but not how to find the rate constant with the given information. Please help/explain. Thanks!

Based on the graphs, the reaction is first order. Because of this, the slope = -k (which is the rate constant). Therefore, to find the rate constant, find the slope of the straight line and take the negative value of it. Remember, rate constants cannot be negative.

So for this problem, to find the slope, do [ln(.0450) - ln(1.2100)]/20000-0 (0 indicates the initial time). You will get -1.65 x 10^-4. This is the slope. Since slope = -k, k= - slope. Plug the slope back in, k = -(-1.65 x 10^-4). You get your answer of 1.65 x 10^-4. Since it is a first order reaction, the units for k are 1/s