What Is the Partial Pressure of Methyl Isonitrile After 12.8 Hours?

[brake4country]

Homework Statement

The rate constant for the conversion of methyl isonitrile is 5 x 10^-5. A scientist has a container containing this substance with a partial pressure of 100 torr. After 12.8 hours (46,000 seconds), what is the partial pressure of methyl isonitrile gas inside the container?

Homework Equations

PV = nRT

The Attempt at a Solution

I tried to attempt this problem using concentrations but they were not given. I have no idea where to start...

 

[Borek]

Partial pressure is directly proportional to the concentration, so as long as there is no change in volume/total pressure they can be used interchangeably (simple conclusion of the Avogadro's law). But I have a different problem - I don't see how to solve, unless you are expected to assume first order reaction.

 

[epenguin]

Yes it sounds you are supposed to assume this. If we'd been told the units of the rate constant that would have told us. Bad! But if the OP knows what the question will have that answered.

We're not even told what the reaction is. Again from the absence of information we'd guess probably just rearrangement to acetonitrile. Again quite plausibly but not necessarily that is just a monomolecular internal movement that could be first order.

 

[brake4country]

Yes, I just realized that the book has this equation: log [A]_t=-kt/2.303 + log[A]₀ for a first order reaction. So this is a first order reaction. This looks like plug and chug:

log[A]_t = -(5x10^-5)(46000)/2.303) + log [100]
log[A] = -1+2 = log[A} = 1 therefore, the answer is 10 torr.

My other question is that I had to use a calculator for this problem but the test does not allow calculators. How best to approach this type of problem without a calculator? Thanks in advance!

 

[epenguin]

Yes, I just realized that the book has this equation: log [A]_t=-kt/2.303 + log[A]₀ for a first order reaction. So this is a first order reaction. This looks like plug and chug:

log[A]_t = -(5x10^-5)(46000)/2.303) + log [100]
log[A] = -1+2 = log[A} = 1 therefore, the answer is 10 torr.

My other question is that I had to use a calculator for this problem but the test does not allow calculators. How best to approach this type of problem without a calculator? Thanks in advance!

In general you can't. It would not be reasonable to ask you do so and it will not happen - unless the problem had nice round numbers like this one. It was reasonable to expect you to know the log of 100 and the antilog of 1.

You would find it kinda useful often if you mastered an art of mentally converting 1st order rate constants to half-lives.

So would I.