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.)

To derive the formula for the rate law, we can start by looking at the balanced chemical equation:

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

The rate law describes how the rate of the reaction (in this case, the rate of decrease in the partial pressure of NO) depends on the concentrations of the reactants. The rate law for this reaction is:

-Δ[NO]/Δt = k[NO]^2[O2]

Where:
- Δ[NO]/Δt is the rate of decrease in the concentration of NO over time.
- k is the rate constant.
- [NO] and [O2] are the concentrations of NO and O2, respectively.

Now, the partial pressure of a gas can be related to its molar concentration using the ideal gas law, which is given by:

PV = nRT

Where:
- P is the pressure of the gas
- V is the volume
- n is the amount of substance (in moles)
- R is the ideal gas constant
- T is the temperature in kelvin

Rearranging the ideal gas law to solve for n/V (molar concentration), we have:

n/V = P/RT

Since the rate law is expressed in terms of concentrations, we need to express the partial pressures of NO and O2 in terms of molar concentrations. Assuming that the volume and temperature are constant, the partial pressure of a gas can be directly related to its molar concentration. Therefore, we can rewrite the rate law in terms of partial pressures:

-Δ(P_NO)/Δt = k([NO]^2)([O2])

Where:
- P_NO is the partial pressure of NO
- P_O2 is the partial pressure of O2

Now, we have the rate law in terms of partial pressures. To find the rate of decrease of the partial pressure of NO in mmHg per second, we can use the given initial pressures of NO and O2 and substitute them into the rate law to calculate the rate.