Question

1.Which acid is the strongest?
  
H2X, Ka1 = 1×10-4
  
HX, Ka = 1×10-7
  
H2X, Ka1 = 1×10-5
  
HX, Ka = 1×10-5

2.Radioactive decay is a first-order rate process. The time required to reduce the activity from 800 counts/min to 100 counts/min for a radioactive sample with a half−life of 30 years is

3.This two-step mechanism has been suggested for a reaction:

N2O(g) + H2(g) N2(g) + H2O(g)

H2(g) + 2NO(g) N2O(g) + H2O(g)

3.Which choice correctly identifies an intermediate and a catalyst?

Intermediate Catalyst

N2O(g) NO(g)

none N2O(g)

N2O(g) none

H2O(g) none

Answer 1

1. Among the given acids, the acid with a high dissociation is H₂X with Ka₁ = 1×10-4

Explanation: An acid with a high dissociation constant is considered a strong acid.

2. The activity of the radioactive sample can be written as shown below.

-dN/dt = k*N

The half life of the radioactive sample = 0.693/k, i.e. 30 years, i.e. 30*365*24*60 min

Therefore, k = 0.693/(30*365*24*60), i.e. 4.395*10^-8 min^-1

Initial activity = 800 counts min^-1

Final activity = 100 counts min^-1

The expression for the rate constant of first order reaction can be written as shown below.

k = (1/t) ln(N₀/N)

Therefore, the time required, t = (1/4.395*10^-8) * ln(800/100), i.e. 90 years.

Note: The answer comes in minutes, which you have to convert into years as shown above.

3. The given reaction sequence can be written in the following way.

2NO(g) + H₂(g) N₂O(g) + H₂O(g)

N₂O(g) + H₂(g) N₂(g) + H₂O(g)

Here, one can easily conclude that N₂O is intermediate, and the catalyst is none.