A 1.9 L reaction vessel, initially at 298 K, contains nitrogen gas at a partial pressure of 337 mmHg and oxygen gas at a partial pressure of 580 mmHg . What is the pressure of N2O4 in the reaction vessel after the reaction?

The pressure of N₂O₄ in the reaction vessel after the reaction is 290 mmHg

Explanation:

Nitrogen gas reacts with oxygen gas to form dinitrogen tetroxide.  

N₂ (g) + 2O₂ (g) →  N₂O₄ (g)

Therefore since by Avogadro's law equal volumes of all gases contain equal numbers of molecules, there fore as the gases are within the same vessel, thier partial pressure is equivalent to their concentration

from the reaction, 1 mole of N₂ react with 2 moles of O₂ to produce 1 mole of N₂O₄

Thus

1 mmHg of N₂ react with 2 mmHg of O₂ to produce 1 mmHg of N₂O₄

337 mmHg N₂ ×(1 mmHg of N₂O₄/ 1 mmHg of N₂) = 337 mmHg N₂O₄

580 mmHg O₂ ×(1 mmHg of N₂O₄/ 2 mmHg of O₂) = 290 mmHg N₂O₄

As seen from the above calculation, the limting reactant is oxygen and the partial pressure of N₂O₄ = 290 mmHg

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years