"SOLUTIONS of Assignment =>1" on "Chemical Kinetics" chapter : 4

These are "SOLUTIONS of  Assignment =>1" on "Chemical Kinetics" chapter : 4 for practice of CBSE BOARD, CBSE NEET, CSIR NET Chemical Science etc.

Topic: -Rate of reaction –Average and instantaneous rates.

Ans.1. The branch of chemistry that deals with the study of reaction rates and their mechanisms is called chemical Kinetics.

Ans.2. Rate of reaction can be defined as the change in concentration of a reactant or product per unit time.

Ans.3. Average rate of a reaction is defined as the change in concentration of a reactant or a product per unit time. It can be determined by dividing the change in concentration of reactant or product by the time interval

For the reaction: A ==> B

R_AV = - Δ[A] / Δt =  + Δ[B] / Δt

Ans.4. The units of rate of a reaction are Mol L^-1S^-1.

Ans.5. The rate of a reaction at a particular moment of time is called instantaneous rate of a reaction.

For a reaction: A ==> B

R_AV = - d[A] / dt =  + d[B] / dt

Where dt = the smallest possible time interval (Δt => 0)

Ans.6.   (i)  In terms of reactants:
              R₁ = - ( 1/4 ) ( Δ[NH₃] / Δt )                       R₂ = - ( 1/5 ) ( Δ[O₂] / Δt )

 In terms of Products:
              R₃ =  + ( 1/4 ) ( Δ[NO] / Δt )                       R₄ = + ( 1/6 ) ( Δ[H₂O] / Δt )

1/4 R₁ = 1/5 R₂ = 1/4 R₃ = 1/6 R₄

- ( 1/4 ) ( Δ[NH₃] / Δt )   =  - ( 1/5 ) ( Δ[O₃] / Δt )
=   + ( 1/4 ) ( Δ[NO] / Δt ) = + ( 1/6 ) ( Δ[H₃O] / Δt )

(ii) In terms of Reactants:
R₁ =  - ( Δ[N₂O₅] / Δt )

In terms of Products:
R₂ =  + ( Δ[NO₂] / Δt )                      R₃ =  + ( Δ[O₂] / Δt )

1/2 R₁ = 1/2 R₂ = R₃

- ( Δ[N₂O₅] / Δt ) =  + ( Δ[NO₂] / Δt )     =   + ( Δ[O₂] / Δt )

Ans.7. Rate of reaction = change in concentration / time interval

= [ (0.8420 0.2105) mol/L]    /    2 hr

= 0.6315 / 2

= 0.3158 mol /L/hr

Ans.8. While writing the expression for rate of a reaction is terms of reactants, there is a negative sign which indicates a decrease in concentration of reactants with time.

Ans.9. From the equation 2O₃ (g) ⇌ 3O₂ (g)

-1/2 ( Δ[O₂] / Δt ) = -3/2 ( Δ[O₃] / Δt ) = -3/2 x (-5.0 x 10^-4 atms^-1) = 7.5 x atms ^-1

Ans.10. 2A ⇌ 4B + C

- 1/2 ( d[A] / Δt ) = 1/4( d[B] / Δt ) = ( d[C] / Δt )

i) Rate of disappearance of B = ( 5 x 10^-3) / 10⁵   mol/ L^-1

= 5 x 10^-4 mol L^-1 s^-1

ii) - d[A] / dt =  2 / 4 d[B] / dt = 1 / 2  d[C] / dt

   

    =  1/2 x 5 x 10^-4 mol L^-1s^-1 

    = 2.5 x 10⁴ mol L^-1 s^-1