Rate Law - Isolation Method

[cvc121]

Homework Statement

Due to formatting problems, I have attached the problem below.

Homework Equations

The Attempt at a Solution

Plotting the appropriate graphs and based on linearity, I have determined that the reaction is first order with respect to [O₂] and second order with respect to [NO]. Therefore, the reaction is third order overall and this corresponds to M^-2s^-1 as the units for the rate constant, k.

However, I am unsure of how to solve for the rate constant. I know that this is an isolation method problem where the initial concentration of one reactant is much smaller than the initial concentration of the others. As such, the large concentrations barely change during the reaction and are approximately constant.

I also know that I will be needing the slopes of the graphs. For the graph of [O₂] in experiment 1, the slope is 0.2328. The slope of the graph of [NO] for the second experiment is 0.0000238.

For instance, in experiment 1, the concentration of NO would stay approximately constant throughout the reaction because it is so large relative to the concentration of [O₂]. So I could write the rate constant as R = k_obs (observed rate constant - corresponds to slope of the graph) [O₂].

Do I just plug in the numbers? Using experiment 1:
k = observed rate constant / [O₂] = 0.2328 / 4.10 x 10^-4 = 567.8 M^-2s^-1?

I do not think this is correct because going through the same process for experiment 2, I get a different result.

Can anyone provide some guidance? Thanks. All help is very much appreciated.

 

[mjc123]

Compare your equation R = k_obs[O₂] with the true rate equation. What is k_obs equal to?

 

[cvc121]

Compare your equation R = k_obs[O₂] with the true rate equation. What is k_obs equal to?

My true rate equation is R = k [NO]²[O₂]. Therefore, the observed rate constant, k_obs, is equal to k [NO]². Rearranging this, I get k = k_obs / [NO]². So do I plug the values from experiment 1 into this equation to get the actual rate constant?

k = 0.2328 / [9.63 x 10^-1]² = 0.251 M^-2s^-1?

 

[mjc123]

Should be. How does it compare with what you get from experiment 2? At least it has the right units, which it would have been obvious your previous attempt didn't, if you had included the units rather than just writing down numbers. I can't emphasise enough the importance of ALWAYS THINKING UNITS, it will save you a lot of trouble and mistakes.

 

[Chestermiller]

Please show your calculations, and particularly, your calculations of the slopes. Your approach and mjc123's comments are correct. There must be an arithmetic error or units error somewhere. It's hard to help without seeing more details of the calculations (particularly for the 2nd order NO case).

 

[epenguin]

You keep saying 'the slopes' but without saying the slopes of what against what I think.

My first reaction seeing the question was what clever experimenters these homework setters are! They chose their two concentrations and timed their measurements exactly so that at each time the concentration was half of what it was the last time! I expect they did this to enable you to see at first glance the reaction is first order in [O₂] which was too obvious for you to mention, but it's not as easy as they make it seem to set up an experiment like that.