Calculating O2 Volume from First-Order Decomposition of Hydrogen Peroxide

[needphyshelp]

Homework Statement

Hydrogen peroxide undergoes a first-order decomposition to water and O2 in aqueous solution. The rate constant at 25°C is 7.40e^-4s. Calculate the volume of O2 obtained from the decomposition reaction of 1.00 mol H2O2 at 25°C and 740 mmHg after 12.4 min.

Homework Equations

ln[H₂O₂]= -kt + ln[H₂O₂]₀

The Attempt at a Solution

12.4 minutes = 744s
k = 7.4e-2

I started using the above equation to calculate the concentration of H2O2 and then tried to get O2 from there, but that just gave me molarity and I'm not sure how to find volume. Also, the equation does not include the pressure, so I am not sure if I am using the correct equation, but I can't find anything else in the book.

Thanks :-)

 

[Borek]

How much oxygen produced per mole of peroxide decomposed? Look at the reaction equation.

Pressure is needed only for ideal gas equation here.

 

[needphyshelp]

1/2 mole of oxygen is produced her mole of peroxide decomposed, but I'm not sure I understand how that helps me.

 

[Borek]

How much peroxide will be decomposed during 12.4 minutes?

 

[needphyshelp]

I think 1.23e-24 moles of H2O2 will decompose, which means 6.14e-25 moles of O2 will be produced, but I still need to find volume. Can I use the ideal gas law to find this?

 

[Borek]

You are on the right track now whe it comes to the approach to find the volume, but your numbers (in the range of 10^-24 moles) are for sure wrong.

 

[needphyshelp]

wow. i just recalculate that and have no idea how i got that number.
does .577 moles of H2O2 decomposing sound more reasonable? and then .289 O2 formed?

using ideal gas law, PV=nRT I have (740)V=.289(62.36)(298.15). V=7.26, but that is not the right answer...
im lost.

 

[Borek]

Looks to me like you are in the correct range with these numbers.

Check whether 0.577 is decomposed - or left, that's most likely mistake. I am leaving for an hour or so now, so you are on your own.

 

[needphyshelp]

Oops. That was the amount left, not decomposed.

Thanks so much for your help! I really appreciate it!