Chemical Kinetics

NCERT InText Questions

Question 1: For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.




`=4xx10^(-4) M mi\n^(-1)``=6.66`

`=6.66xx10^(-6) M s^(-1)`

Question 2: In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol L-1 in 10 minutes. Calculate the rate during this interval.


`R_(av)=(-1Δ[A])/(2 Δt)=-1/2xx([A]_2-[A]_1)/(t_2-t_1)`


`=5xx10^(-3) M mi\n^(-1)`

Question 3: For a reaction A + B → Product, the rate law is given by `r=k[A]^(1/2)[B]^2` What is the order of this reaction?

Answer: Order of reaction `=1/2+2=(1+4)/2=5/2=2.5`

Question 4: The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Answer: The reaction can be written as: `X→Y`

As per rate law,

Rate `=k[X]^2`

If [X] is increased 3 times, then rate can be given as follows:

Rate `=k[3X]^2`

`=9k[X]^2=9 ra\te`

So, the rate increases by 9 times.

Question 5: A first order reaction has a rate constant 1.15 × 10-3 s-1. How long will 5 g of this reactatn take to reduce to 3 g?

Answer: Given [A]0 = 5g, [A] = 3 g and `k=1.15xx10^(-3)s^(-1)`

For 1st order reaction,



`=2.00xx10^3(lo\g 1.667)`

`=443.8  s`

Question 6: Time required to decompose SO2Cl2 to half of its initial amount is 6.0 minutes. If the decomposition is a first order reaciton, calculate the rate constant of the reaction.

Answer: For 1st order reaction:




Question 7: What will be the effect of temperature on rate constant?

Answer: The rate of most of the chemical reactions increases with increase in temperature. The rate constant nearly doubles with rise in temperature by 10° for a chemical reaction.

Question 8: The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea.

Answer: Given, T1 = 298 K, T2 = 308 K, k1 = k, and k2 = 2k

We know,


Or, `lo\g\(2k)/k=(E_a)/(2.303xx8.314)((308-298)/(308xx298))`

Or, `lo\g\ 2=(E_a)/(2.303xx8.314)xx(10)/(308xx298)`

Or, `E_a=((log\ 2)(2.303xx8.314)(308xx298))/(10)`

`=52897.7 J mo\l^(-1)`

`=52.8 kJ mo\l^(-1)`

Question 9: The activation energy for the reaction


Is 209.5 kJ mol-1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

Answer: Fraction of molecule having energy equal to or greater than activation energy can be calculated as follows:


So, `In x=(-E_a)/(RT)`

Or, `lo\g x=-(209.5xx10^3)/(2.303xx8.314xx581)``=-18.8323`

So, `x=an\ti\lo\g\(-18.8323)=1.471xx10^(-19)`

Copyright © excellup 2014