Question

The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-1. If the initial pressure of N2O is 5.10 atm at 730°C, calculate the total gas pressure after one half-life. Assume that the volume remains constant.

Answer 1

Answer:

The total pressure after one half is 6.375 atm.

Explanation:

The initial pressure of product is increases while the pressure of reactant would decrease.

Balanced chemical equation:

2N₂O  → 2N₂ + O₂

The pressure of N₂O is 5.10 atm. The change in pressure would be,

N₂O = -2x

N₂ = +2x

O₂ = +x

The total pressure will be

P(total) = P(N₂O) + P(N₂)  + P(O₂)

P(total) = ( 5.10 - 2x) + (2x)  + (x)

P(total) = 5.10 + x

After one half life:

P(N₂O)  = 1/2(5.10) = 5.10 - 2x

x = 5.10 - 1/2(5.10) /2

x = 5.10 - 0.5 (5.10) /2

x = 5.10 - 2.55 / 2

x = 2.55 /2 = 1.275 atm

Thus the total pressure will be,

P(total) = 5.10 + x

P(total) = 5.10 + 1.275

P(total) = 6.375 atm