This reaction was monitored as a function of time:
AB → A + B
A p...

Question

This reaction was monitored as a function of time:
AB → A + B
A plot of 1/[AB] versus time yields a straight line with a slope of
+0.55/Ms.
b. Write the rate law for the reaction.

Accepted Answer

Hey everyone in this example, we need to write the rate law for the reaction of X producing Y. We're told that the plot of one over X. One over our concentration of X versus time yielded the following slope. So because according to this prompt, we specifically have a plot of one over our concentration of X versus time, where X. We recognize is our reactant. This sort of plot tells us that we should have a linear rate law that would therefore be second order. So we should recall that our second order integrated rate law is one over the concentration of our reactant at a given time is equal to the rate constant times time plus one over our concentration of our reactant initially. So this is our integrated second order rate law. And we can relate this directly. Like in this example to the equation of a line where we have why? Sorry? Where we have Y representing one over our concentration of our reactant at a given time equal to M X. Which represents the place of the rate constant K. Times time. So just to be clear, K would represent m our slope where K we recall is our rate constant. And then we would have plus our Y intercept B, which represents one over our initial concentration of our reactant. So because we recognize from the equation of our line and from our integrated second order rate law, that slope is related to our rate constant K. And according to the prompt, we have a slope of positive 72 or positive 0. polarities times inverse seconds. We can say that therefore our rate law for the reaction of ex producing Y is equal to Our slope which is again given in the prompt as point positive . inverse polarity times in verse. Second since we're moving them to the numerator, which is then multiplied by our concentration of our reactant, which from the prompt is X. And because this is a second order rate law, we're going to raise this to an exponent of two. And so this would be our final answer for our second order rate law based on our equation of our line that is generated from our plot of one over concentration of reacting X versus time. So everything highlighted in yellow is our rate law expression. If you have any questions, please leave them down below. And I will see everyone in the next practice video.

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