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 your post below.
To derive the formula needed to calculate the rate of decrease of the partial pressure of NO, we can follow these steps:
Step 1: Write the given rate law equation:
-Δ[NO]/Δt = k[NO]^2[O2]
Step 2: Convert the partial pressures to molar concentrations:
From the ideal gas law, we know that the molar concentration (mol/L) of a gas is related to its partial pressure (in atm) by the equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature (in Kelvin).
Rearranging the ideal gas law equation, we get: n/V = P/RT
Since we are interested in the molar concentration (n/V) of NO, we can write it as: [NO] = P(NO) / RT, where P(NO) is the partial pressure of NO.
Similarly, for O2, we have: [O2] = P(O2) / RT, where P(O2) is the partial pressure of O2.
Step 3: Substitute the molar concentrations into the rate law equation:
-Δ[NO] / Δt = k[NO]^2[O2]
Substituting [NO] and [O2] from step 2, we get:
-Δ(P(NO) / RT) / Δt = k(P(NO) / RT)^2 (P(O2) / RT)
Simplifying the equation further:
-1 / RΔt ∙ Δ(P(NO) / T) = kP(NO)^2 (P(O2) / RT)
Step 4: Rearrange the equation to get the rate of decrease of the partial pressure of NO:
To find the rate of decrease of the partial pressure of NO, we need to isolate ΔP(NO) / Δt.
Rearranging the equation:
Δ(P(NO) / T) / Δt = -k[P(NO)^2(P(O2) / RT)] ∙ R
Since we are interested in the rate of decrease, we can take the negative sign inside:
-Δ(P(NO) / T) / Δt = k[P(NO)^2(P(O2) / RT)] ∙ R
The negative sign indicates that the partial pressure of NO is decreasing.
Step 5: Calculate the rate of decrease of the partial pressure of NO:
Now that we have the formula -Δ(P(NO) / T) / Δt = k[P(NO)^2(P(O2) / RT)] ∙ R, we can substitute the given values to find the rate of decrease of the partial pressure of NO.
Given values:
k = 1.16 x 10^(-5) M^(-2)∙s^(-1) (rate constant)
P(NO) = 155 mmHg (initial partial pressure of NO)
P(O2) = 345 mmHg (initial partial pressure of O2)
R = 0.0821 L∙atm/(mol∙K) (gas constant)
T = 339 °C = 339 + 273 = 612 K (temperature in Kelvin)
Substituting the values, we get:
rate of decrease of the partial pressure of NO = -[(-1.16 x 10^(-5) M^(-2)∙s^(-1))(155 mmHg)^2 (345 mmHg) ∙ R] / (0.0821 L∙atm/(mol∙K) ∙ 612 K)