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 3.50 atm at 730°C, calculate the total gas pressure after one half-life. Assume that the volume remains constant. Insert your answer in decimal notation rounded to 3 significant figures.

 the total gas pressure after one half-life is 4.38 atm

Explanation:

The balanced equation for the decomposition of N2O to N2 and O2 is given as:

2N₂O(gas) ⇒ 2N₂(gas) + O₂(gas)

2 moles of N₂O produce 2 moles of N₂ and 1 mole of O₂

The change in pressure depends on the coefficient (number of moles) of the reactant and product.

                                                          N₂O                    N₂                     O₂

number of moles                                2                        2                        1

Initial pressure (atm)                           3.50                  0                        0

change in pressure                            -2x                    +2x                      x

Final pressure (atm)                            3.50 - 2x           2x                       x

The total final pressure is the sum of the individual total pressure. i.e.:

Total final pressure = final pressure of N₂O + final pressure of N₂ + final pressure of O₂

Total final pressure = (3.5 - 2x) + (2x) + x

Total final pressure = 3.5 + x

After one half life, the initial pressure of N₂O would be half its value.

Final pressure of N₂O = half of the initial pressure of N₂O

3.5 - 2x = 0.5(3.5)

3.5 - 2x = 1.75

2x = 1.75

x = 0.875 atm

Therefore, Total final pressure = 3.5 + x = 3.5 + 0.875

Total final pressure = 4.38 atm to 3 significant figures

the answer is true. hope it .

the answer is 0.8m.

by: nicholas