question in the description. I know that partial pressure of NO=O

Question

Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.650 atm, and the small bulb, with a volume of 1.50 L, contains oxygen at a pressure of 2.50 atm. The temperature at the beginning and the end of the experiment is 22∘C. After the stopcock is opened, the gases mix and react. 2NO(g)+O2(g)→2NO2(g) Which gases are present at the end of the experiment? NO O2 NO2 What are the partial pressures of the gases? If the gas was consumed completely, put 0 for the answer. 𝑃NO= ______atm 𝑃O2=______atm 𝑃NO2=______atm.

Answer 1

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

Approach: find moles of NO in large bulb, and find moles O2 in the small bulb by using the ideal gas law:

PV = nRT and n = PV/RT

moles NO = (0.650 atm)(6.00 L) / (0.0821 Latm/Kmol)(295K) = 0.161 mols NO

moles O2 = (2.50 atm)(1.50 L) / (0.0821 Latm/Kmol)(295K) = 0.155 mols O2

Next, determine which reactant is limiting:

From the mol ratio in the balanced equation, we see that NO is limiting as it takes 2x the mols of O2.

Set up an ICE table to determine moles of each gas at equilibrium:

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

0.161......0.155.............0..............Initial

-0.161....-0.0805..........+0.161......Change

0........0.0745...........0.161........Equilibrium

These values are the moles of each gas present at equilibrium. To convert to partial pressure, we first find total volume (7.50 L) and and then use ideal gas law to determine pressure of each:

PV = nRT and P = nRT/V

Pressure of O2 = (0.0745 mol)(0.0821 Latm/Kmol)(295K) / 7.50 L = 0.241 atm

Pressure of NO2 = (0.161 mol)(0.0821 Latm/Kmol)(295K) / 7/50 L = 0.520 atm

Pressure of NO = 0 atm, as all of it is consumed in the reaction