Question

Im studying for the MCAT and i found this question in my old chem book. i can't figure out how to derive the formula they are talking about...

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide.

2 NO(g)+ O2(g) --> 2 NO2(g)


The rate law is -[NO]/t = k[NO]2
[O2], where the rate constant is 1.16 x 10-5 M^-2∙s^-1at 339oC.

The initial pressures of NO and O2 are 155 mmHg and 345 mmHg, respectively. What is the rate of decrease of partial pressure of NO in mmHg per second?

(Hint: From the ideal gas law, obtain an expression for the molar concentration (mol/L) of a particular gas in terms of its partial
pressure.)

Answer 1

Could this be as follows:
PV = nRT
with P in atm, V in L, n = mols and R and T usual.'
P = n(RT)/V
Since n/V = mols/L that is M so
p = MRT.

Answer 2

where does the rate law and rate constant come into play?

Answer 3

Your post said you didn't know how to derive the formula. I did that for you. Do you have an answer for the problem? I assume so since this is a study guide.

Answer 4

Book doesn't have an answer for this one.

Im just lost as to how to apply the ideal gas law to the rate law and constant.

P=MRT => 155atm=(2mol)(0.0821)(298k)?

Answer 5

P is 155/760 = ? if you are using R = 0.08205 and T is 273+339 = ?
That gives you (NO)in mols/L and you plug that into the rate law to calculate [delta(NO)/delta T] and convert that back to mm Hg using P = MRT (remember you will get atm and atm x 760 = mm Hg)