This reaction was monitored as a function of time:
A → B + C
A pl...

Question

This reaction was monitored as a function of time:
A → B + C
A plot of ln[A] versus time yields a straight line with
slope -0.0045/s.
a. What is the value of the rate constant (k) for this reaction at
this temperature?

Accepted Answer

everyone in this example, we need to identify the rate constant K. For the reaction of X producing Y. We're told that the plot of our natural log of our concentration of X versus time Yielded a straight line with a slope of negative 0. per second. So what we're going to do is recall that our first order integrated rate law which we know we're following because according to the prompt we have the concentration of our reactant X versus time. We would recall that this rate law is found from taking the natural log of our concentration of our reactant at a given time equal to negative one times our great constant K. Times time in seconds added to the natural log of our concentration of our reactant initially. And we're relating this to the equation of a line where we have y equal to M X plus B. Where we would recall that the m portion here corresponds to our slope which is equivalent to negative one times our rate constant K. So if we know that our slope is equal two -K. We can say that Therefore negative one times our slope will give us our value for our rate constant K. And so we can say that our rate constant K. Is equal to negative one times our Slope given in the prompt as negative 0.0079 per second. And we would see that because we have a negative times a negative. This would give us actually a positive value for a rate constant equal to 0.0079/s. So this would be our final answer here for our value of our rate constant K. So I hope that everything I explained was clear. If you have any questions, just leave them down below. Otherwise I'll see everyone in the next practice video.

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