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.)

See below

To derive the formula for the rate of decrease of partial pressure of NO, we need to understand the relationship between partial pressure and molar concentration.

According to the ideal gas law, the relationship between pressure (P), volume (V), and the number of moles (n) of a gas is given by the equation: PV = nRT, where R is the ideal gas constant and T is the temperature in Kelvin.

Rearranging the ideal gas law, we get: P = (n/V)RT

Now, to express the molar concentration (C) of a gas in terms of its partial pressure (P), we divide both sides of the equation by the volume (V), resulting in: C = (n/V) = P/RT

Using this expression for molar concentration, let's continue to derive the formula for the rate of decrease of the partial pressure of NO.

Given the rate law: -Δ[NO]/Δt = k[NO]^2[O2]

The rate of decrease of the partial pressure of NO (-ΔP(NO)/Δt) is directly proportional to the rate of change of the molar concentration of NO, which can be expressed as -ΔC(NO)/Δt.

Substituting the expression for molar concentration (C) into the rate law, we have:

-ΔC(NO)/Δt = k(C(NO))^2(C(O2))

Now, recall that the partial pressure (P) of a gas is related to its molar concentration (C) by P = CRT. We can substitute this relationship into the equation:

-Δ(P(NO)/RT)/Δt = k((P(NO)/RT))^2(C(O2))

Rearranging the equation, we get:

-ΔP(NO)/Δt = (k/R) * ((P(NO))^2 / (RT)) * (C(O2))

Since R and T are constant, let's combine them with the rate constant (k/R) into a new constant, k'.

Therefore, the derived formula for the rate of decrease of the partial pressure of NO is:

-ΔP(NO)/Δt = k' * (P(NO))^2 * (C(O2))

To calculate the rate of decrease of the partial pressure of NO in mmHg per second, substitute the given values: P(NO) = 155 mmHg and P(O2) = 345 mmHg, into the formula.