Nitric Oxide reacts with oxygen as follows:

2NO + O2  2NO2

The rate law is –[NO]/t = k[NO]2 [O2], where k is 1.16x10-5 L 2 /(mol2 s) at 339C. A vessel contains NO and O2 at 339 C. The initial partial pressure 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 of a particular gas in terms of its partial pressure.

To determine the rate of decrease of the partial pressure of nitric oxide (NO) in the given reaction, we can start by using the rate law provided and the ideal gas law to relate the partial pressures to molar concentrations. Let’s break this down step by step.

Understanding the Reaction and Rate Law

The reaction we are considering is:

2NO + O2 → 2NO2

The rate law for this reaction is given as:

-Δ[NO]/Δt = k[NO]^2[O2]

Where:

• k is the rate constant (1.16 x 10^-5 L²/(mol²s))
• [NO] is the molar concentration of NO
• [O2] is the molar concentration of O2

Using the Ideal Gas Law

The ideal gas law states that:

PV = nRT

Where:

• P is the pressure of the gas
• V is the volume
• n is the number of moles
• R is the ideal gas constant (0.0821 L·atm/(K·mol))
• T is the temperature in Kelvin

From this, we can express the molar concentration [C] of a gas in terms of its partial pressure (P) as follows:

[C] = P / (RT)

Calculating Molar Concentrations

First, we need to convert the temperature from Celsius to Kelvin:

T = 339 + 273.15 = 612.15 K

Next, we can calculate the molar concentrations of NO and O2 using their initial partial pressures:

For NO:

[NO] = P_NO / (RT) = 155 mmHg / (0.0821 L·atm/(K·mol) * 612.15 K)

We need to convert mmHg to atm (1 atm = 760 mmHg):

P_NO = 155 mmHg * (1 atm / 760 mmHg) = 0.204 atm

Now substituting into the equation:

[NO] = 0.204 atm / (0.0821 * 612.15) = 0.00405 mol/L

For O2:

[O2] = P_O2 / (RT) = 345 mmHg / (0.0821 L·atm/(K·mol) * 612.15 K)

Converting mmHg to atm:

P_O2 = 345 mmHg * (1 atm / 760 mmHg) = 0.454 atm

Now substituting into the equation:

[O2] = 0.454 atm / (0.0821 * 612.15) = 0.00887 mol/L

Substituting into the Rate Law

Now that we have the concentrations, we can substitute them into the rate law:

-Δ[NO]/Δt = k[NO]^2[O2]

Substituting the values:

-Δ[NO]/Δt = (1.16 x 10^-5 L²/(mol²s)) * (0.00405 mol/L)² * (0.00887 mol/L)

Calculating this gives:

-Δ[NO]/Δt = (1.16 x 10^-5) * (0.0000164) * (0.00887) = 2.06 x 10^-10 mol/L/s

Converting to Partial Pressure Decrease

To find the rate of decrease of partial pressure of NO in mmHg/s, we can use the relationship between concentration and pressure:

ΔP/Δt = Δ[NO] * RT

Substituting the values:

ΔP/Δt = (2.06 x 10^-10 mol/L/s) * (0.0821 L·atm/(K·mol)) * (612.15 K)

Calculating this gives:

ΔP/Δt = 1.02 x 10^-8 atm/s

Finally, converting atm to mmHg:

ΔP/Δt = 1.02 x 10^-8 atm/s * 760 mmHg/atm = 7.77 x 10^-6 mmHg/s

Thus, the rate of decrease of the partial pressure of NO is approximately 7.77 x 10^-6 mmHg/s.