Question

Consider this reaction: →2NH3g+N2g3H2g At a certain temperature it obeys this rate law: rate =2.11·M−1s−1NH32 Suppose a vessel contains NH3 at a concentration of 0.590M. Calculate how long it takes for the concentration of NH3 to decrease by 88.0%. You may assume no other reaction is important. Round your answer to 2 significant digits

Answer 1

5.9 s

Step-by-step explanation:

Let's consider the following reaction.

2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)

The rate law is:

rate = 2.11 M⁻¹s⁻¹[NH₃]²

where

2.11 M⁻¹s⁻¹ is the rate constant (k)

2 is the order of reaction for NH₃

The initial concentration of NH₃ ([NH₃]₀) is 0.590 M. If it decreases by 88.0%, the final concentration ([NH₃]) will be 100.0% - 88.0% = 12.0% of the initial concentration, that is, 0.0708 M.

For a second order reaction we can use the following expression.

`(1)/([NH_(3)]) =(1)/([NH_(3)]_(0)) + k.t\\(1)/(0.0708M) =(1)/(0.590M) +2.11M^(-1) s^(-1) .t\\t=5.89s`