Chemical Kinetics: determining order of reaction

[BrettJimison]

Homework Statement

Good day! I have a question from my chem HW that is stumping me. Although the answer may be simple, it has eluded me for quite a bit. It regards rate laws and constants of reactions.

Question is as stated: This reaction was monitored as a function of time: AB --> A + B

A plot of 1/[AB] versus time yields a straight line
with slope= - 0.055 M^-1s^-1

What is the value of the rate constant (k) at this temp?

Homework Equations

SINCE it stated 1/[AB] vs. time, its implying an order 2 reaction since order 2 reaction
has the integrated rate law:
1/[AB]_t = kt + 1/[AB]₀

y-axis = 1/[AB]...

The Attempt at a Solution

When a order 2 reaction is plotted against time, k = slope. When the question stated it was talking about a plot of 1/[AB] its referring to a order 2 reaction, and since order 2 reactions (when plotted against time) have a y-axis = 1/[AB], I am deducing the reaction is order 2.

I'm confused since the given slope is negative, and every graph of an order 2 reaction vs. time in my book has a straight line with a positive slope.

Like I said, its an easy problem, with an easy solution, but I'm not seeing it.

 

[Chestermiller]

Homework Statement

Good day! I have a question from my chem HW that is stumping me. Although the answer may be simple, it has eluded me for quite a bit. It regards rate laws and constants of reactions.

Question is as stated: This reaction was monitored as a function of time: AB --> A + B

A plot of 1/[AB] versus time yields a straight line
with slope= - 0.055 M^-1s^-1

What is the value of the rate constant (k) at this temp?

Homework Equations

SINCE it stated 1/[AB] vs. time, its implying an order 2 reaction since order 2 reaction
has the integrated rate law:
1/[AB]_t = kt + 1/[AB]₀

y-axis = 1/[AB]...

The Attempt at a Solution

When a order 2 reaction is plotted against time, k = slope. When the question stated it was talking about a plot of 1/[AB] its referring to a order 2 reaction, and since order 2 reactions (when plotted against time) have a y-axis = 1/[AB], I am deducing the reaction is order 2.

I'm confused since the given slope is negative, and every graph of an order 2 reaction vs. time in my book has a straight line with a positive slope.

Like I said, its an easy problem, with an easy solution, but I'm not seeing it.

If [AB] is decreasing with time, then 1/[AB] must be increasing with time. That's why the slope is positive.

Chet

 

[BrettJimison]

Thanks for the reply, but 1/[AB] has a negative slope not positive in this problem. That's why I'm confused. The graph would indicate that 1/[AB] is decreasing with time which means [AB] is increasing which makes no sense.

 

[Chestermiller]

Thanks for the reply, but 1/[AB] has a negative slope not positive in this problem. That's why I'm confused. The graph would indicate that 1/[AB] is decreasing with time which means [AB] is increasing which makes no sense.

Oh. I didn't notice that. It must just be a mistake in the problem statement.

Chet

 

[DeeNos]

Dear student,

Thank you for reaching out with your question about chemical kinetics and determining the order of reaction. It's great that you are working on this problem and trying to understand it better.

Firstly, you are correct in your understanding that the given plot of 1/[AB] versus time implies an order 2 reaction. As you mentioned, the integrated rate law for a second-order reaction is 1/[AB]t = kt + 1/[AB]0, where [AB]t is the concentration of AB at time t, k is the rate constant, and [AB]0 is the initial concentration of AB.

Now, let's address your confusion about the negative slope. The slope of a plot of 1/[AB] versus time for a second-order reaction is indeed positive. However, in this case, the negative slope indicates that the concentration of AB is decreasing with time. This is because as the reaction progresses, AB is being consumed and converted into A and B, leading to a decrease in its concentration.

To determine the value of the rate constant at this temperature, you can use the slope of the plot, which is -0.055 M-1s-1. The negative sign indicates that the reaction is proceeding in the direction of the reactants being consumed, which is also consistent with the given reaction equation AB --> A + B. Therefore, the rate constant (k) at this temperature is 0.055 M-1s-1.

I hope this helps you understand the problem better. Keep up the good work in your studies of chemical kinetics!

Best regards,


Chemist