This graph shows a plot of the rate of a reaction versus the concentration of the reactant.
c. Write a rate law for the reaction including the value of k.

Question

This graph shows a plot of the rate of a reaction versus the concentration of the reactant.

c. Write a rate law for the reaction including the value of k.

Accepted Answer

Hey everyone in this example we're given the plot for the rate versus our concentration of our reactant A. For the reaction of a producing B. And we need to provide the complete rate law for our reaction. So what we should recognize is that because according to the prompt we have rate versus our concentration of A. We should recognize that this is for a first order reaction. And because this is a straight line here, we can say therefore the reaction is definitely a first order reaction with respect to our react in A. So we can write out our rate law for a first order reaction where we have rate equal to the rate constant K. Multiplied by the concentration of our react in A. And we can raise it to the first order. However, we can just leave it blank since we know that this is to a power of one. So our next step is to find what our rate constant K. Is. So we would say that K. Is equal to our slope and recall that our slope can be calculated by taking y two minus Y one over X two minus x one. So we're going to calculate based on our values in the given graph. And we would say that Our slope K. is equal to and our numerator, we have Y two. So we can find Y two by looking at the end of our graph and seeing the Y value that corresponds which is 20.1 to 0 molar per second. So we'll plug that in 0.1 to zero polarity per second subtracted from Y one which according to our prompt comes from the beginning of our graph. And we can say that this value here is about 0.30 polarity per second. So we'll plug that in 0.30 polarities per second in our denominator. We have X two. So X two comes from the end of our graph here where we see we have a value of 1.2 molar. So we would say 1.2 molar subtracted from X one where we see comes from the beginning of our graph being zero moller. So 0.0 moller. And so what we would simplify this too is that our rate constant K. Is equal to a value of 0.75. And as far as units were going to cancel out our moller units here. Leaving us with inverse seconds in the new meters or unit here is inverse seconds. So now that we know the value for our rate constant, we can write out our complete rate law and say that therefore our rate is equal to our rate constant K. Which we now know is 0.75, inverse seconds, multiplied by the concentration of our react in a And so to complete this example as our final answer, we have here, What we have here highlighted in yellow. So I hope that everything I explained was clear. But if you have any questions, just leave them down below. Otherwise, I will see everyone in the next practice video.

This graph shows a plot of the rate of a reaction versus the concentration of the reactant.
c. Write a rate law for the reaction including the value of k.

Verified Solution

Key Concepts

Rate Law

Rate Constant (k)

Reaction Order

This graph shows a plot of the rate of a reaction versus the concentration of the reactant.c. Write a rate law for the reaction including the value of k.

Verified Solution

Key Concepts

Rate Law

Rate Constant (k)

Reaction Order

This graph shows a plot of the rate of a reaction versus the concentration of the reactant.
c. Write a rate law for the reaction including the value of k.