JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Chemical Kinetics

The decomposition reaction

2 N_{2} O_{5} (g) Δ 2 N_{2} O_{4} (g) + O_{2} (g)

is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After

Y \times 1 0^{3} s

, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is

5 \times 1 0^{- 4} s^{- 1}

, assuming ideal gas behavior, the value of Y is ____

Answer: 2.3

Solution

For the reaction

2 N_{2} O_{5} (g) \to 2 N_{2} O_{4} (g) + O_{2} (g)

at constant V and T:

Initial pressure:

P_{0} = 1

atm

At time

t = Y \times 1 0^{3}

s, let pressure change be

P

Total pressure:

P_{T} = (1 - 2 P) + 2 P + P = 1 + P = 1.45

P = 0.45

atm

For first-order kinetics:

(2 k) t = 2.303 lo g (\frac{1}{1 - 2 P})

(2 \times 5 \times 1 0^{- 4}) \times Y \times 1 0^{3} = 2.303 lo g (\frac{1}{0.1})

Y = \frac{2.303 \times 1}{1 \times 1 0 ^{- 3} \times 1 0 ^{3}} = 2.303

Rounded to two decimal places: Y = 2.30

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Numerical answer