[Chemistry] Calculating rate of change (1 Viewer)

[Tecumseh]

These are driving me crazy and I have a quiz tomorrow. Can someone explain this (in a way that is extremely dumbed down)?

Consider the reaction:

2N2O----> 2N2 + O2

In the first 12.0 seconds of the reaction, 1.6×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.260L. What is the average rate of the reaction over this time interval?

Predict the rate of change in the concentration of N2O over this time interval.


. What is the average rate of the reaction over this time interval?



2N2O(g)→2N2(g)+O2(g)

 

[WookieBooger]

is the answer "Blue"?

 

[MikeIke]

Wow, you're a lot smarter than I am

 

[Ilov80s]

Get the concentration of O2 before (0) and after. That is mol/L. Then find the change from beginning to end, which is basically the concentration you calculated for the end because it started at 0. Then divide it by 12 seconds and I think you will have the rate.

For N2, it has a coefficient of 2, so for every 1 mol of O2, you will get 2 mol of N2 so you can use that to determine N2 concentration, etc.

 

[MC Gas Money]

Look at me thread

 

[inca911]

It's been a while, but Average Rate of Reaction is change in concentration (mol/L) over time.

Rate of change is positive as products are being formed.

For each mol of O2 formed, 2 mol of N2O are consumed, so the rate of change of N20 is double, and the sign is negative as reagents are being consumed.

Fixed math is below, as iluv80s is right that it's [Concentration], not just mol.

 

[Ilov80s]

It's been a while, but Average Rate of Reaction is Change in change in concentration (mol) over time.

1.6 x 10-2 mol / 12 seconds = 0.016 mol / 12 sec = 0.00133 mol/sec (rate of change is positive as products are being formed)

For each mol of O2 formed, 2 mol of N2O are consumed, so the rate of change of N20 is double, and the sign is negative as reagents are being consumed = - 0.00266

Reaction vessel size doesn't matter, and is only there to confuse you into trying to use it.

Shouldn't it be change in concentration (mol/L) and not just change in mols?

 

[inca911]

It's been a while, but Average Rate of Reaction is Change in change in concentration (mol) over time.

1.6 x 10-2 mol / 12 seconds = 0.016 mol / 12 sec = 0.00133 mol/sec (rate of change is positive as products are being formed)

For each mol of O2 formed, 2 mol of N2O are consumed, so the rate of change of N20 is double, and the sign is negative as reagents are being consumed = - 0.00266

Reaction vessel size doesn't matter, and is only there to confuse you into trying to use it.

Shouldn't it be change in concentration (mol/L) and not just change in mols?

I think you are correct, so the Liters would do matter. I told you it's been a long time. (shame)

Divide all the mol values by the Liters to turn into Molarity, then doing the same math should give you the answer.

Rate of Reaction O2 = (0.016 mol O2 / 0.260 L) = 6.15 x 10^-2 M / 12 sec = 5.12 x 10^-3 M/sec

Double it and make it negative for the N2O as it's being consumed = -1.03 x 10^-2 M/sec N2O

Do you have an answer key?

 

[Tecumseh]

It's been a while, but Average Rate of Reaction is Change in change in concentration (mol) over time.

1.6 x 10-2 mol / 12 seconds = 0.016 mol / 12 sec = 0.00133 mol/sec (rate of change is positive as products are being formed)

For each mol of O2 formed, 2 mol of N2O are consumed, so the rate of change of N20 is double, and the sign is negative as reagents are being consumed = - 0.00266

Reaction vessel size doesn't matter, and is only there to confuse you into trying to use it.

Shouldn't it be change in concentration (mol/L) and not just change in mols?

I think you are correct, so the Liters would do matter. I told you it's been a long time. (shame)

Divide all the mol values by the Liters to turn into Molarity, then doing the same math should give you the answer.

Rate of Reaction O2 = (0.016 mol O2 / 0.260 L) = 6.15 x 10^-2 M / 12 sec = 5.12 x 10^-3 M/sec

Double it and make it negative for the N2O as it's being consumed = -1.03 x 10^-2 M/sec N2O

Do you have an answer key?

Those are both correct. Thank you. I didn't think the liters mattered, which was what was killing me.

 

[GTBilly]

I clicked this thread just to reassure myself I've forgotten everything about OAC Chem besides how to make fireworks.

 

[Mr. Pickles]

I've got a paper coming out in ACS Nano that you might like.

 

[Tecumseh]

Not likely. I f'n hate chem. Thank the baby jeebus this the last one I have to take.

ETA: But stay close. Working on 0th, 1st, 2nd order half-life and Arrhenius equations. We need a ickles: signal.