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.)
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.
where does the rate law and rate constant come into play?
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.
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)?
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)
To find the rate of decrease of the partial pressure of NO in mmHg per second, we need to use the given rate law and convert it into a form that relates to partial pressures. Here are the steps to derive the formula:
Step 1: Convert the given initial pressures of NO and O2 into molar concentrations (mol/L):
Using the ideal gas law, we can relate the pressure of a gas to its molar concentration. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
Rearranging the equation to solve for concentration (n/V), we get:
n/V = P/RT
Since we want the molar concentration in mol/L, we divide the pressure in mmHg by 760 to convert it to atm:
[NO] = (155 mmHg / 760 mmHg/atm) / RT = (0.204 atm) / RT
[O2] = (345 mmHg / 760 mmHg/atm) / RT = (0.453 atm) / RT
Step 2: Substitute the molar concentrations into the rate law expression:
-Δ[NO]/Δt = k[NO]^2[O2]
Substituting the molar concentrations, the expression becomes:
-Δ(0.204/RT)/Δt = k(0.204/RT)^2(0.453/RT)
Simplifying further:
-Δ(0.204/RT)/Δt = k(0.204^2/RT^2)(0.453/RT)
Since we are looking for the rate of decrease of NO, we can assume Δ(0.204/RT) ≈ -Δ[NO].
So the expression becomes:
-Δ[NO]/Δt = k(0.204^2/RT^2)(0.453/RT)
Step 3: Simplify the expression and calculate the rate of decrease of partial pressure of NO in mmHg per second:
To find the rate of decrease of partial pressure of NO in mmHg per second, we need to find the rate of decrease of [NO] and convert it back to pressure units (mmHg). Since the rate of decrease is given in terms of Molarity per second (M/s), we multiply it by (RT/0.204) to convert it to partial pressure units (mmHg/s):
Rate of decrease of partial pressure of NO = (-Δ[NO]/Δt) * (RT/0.204)
Plugging in the given values:
Rate of decrease of partial pressure of NO = (-1.16 x 10^-5 M^-2∙s^-1) * (0.0821 L·atm/mol·K) * (339 + 273 K) / (0.204)
Calculating the rate of decrease of partial pressure of NO will give you the final answer in mmHg/s.