Question

2NH3------->N2+3H2
Initially, 3.4g of ammonia is heated in a 5 litre vessel. Calculate the time interval for 40% disappearance of NH3 while rate constant is 2X10^-5 M.sec^-1

Answer 1

NUmbe rof moles of NH₃ = 3.4/17 = 0.2 moles
Initial concentration of NH₃, [NH₃]₀ = 0.2/5 = 0.04 mole/L
40% of 0.04 = 0.04×(40/100) = 0.016 mole/L
after 40% disappearance, [NH₃] = 0.04 - 0.016 = 0.024 mole/L
let the time taken = t

rate constant = 2×10^-5 M.sec^-1 = 2×10^-5 mole L^-1 sec^-1 
from unit of rate constant, it is a zeroth order reaction.

[NH₃] = [NH₃]₀ - kt
⇒0.024 = 0.04 - kt
⇒ 0.024 - 0.04 = - 2×10^-5× t
⇒t = (-0.016)/(-2×10^-5)
⇒ t = 800s