//M2//QN1//SUB//DL0//EQ

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

//X

Average rate =
=
=
=
= 4 × 10–4 M min–1
Rate of reaction in terms of second
=
= 6.67 ×10–6 M s–1

//M2//QN2//SUB//DL0//EQ

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 ?

//X

For reaction 2A → Product,
∆t = 10 min
Rate =
=
= mol L–1 min–1
= 5.0 ×10–3 M min–1

//M1//QN3//SUB//DL0//EQ

For reaction A + B → Product; the rate law is given by, r = k[A][B]2 What is the order of the reaction?

//X

Overall order of reaction is equal to the sum of powers of the concentration of reactants
Here, order of the reaction = + 2 = 2.5

//M2//QN4//SUB//DL0//EQ

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times How will it affect that rate of formation of Y?

//X

X → Y is a second order reaction,
Therefore, rate law can be expressed as
V1 = k[X]2
If the concentration of X is increased three time, then the new concentration will be 3X and the new rate will be
V2 = k[3X]2 = 9k[X]2
Taking ratio of V2 and V1, we get
V2 = 9V1
Hence the rate of formation of Y will increase by 9 times.

//M3//QN5//SUB//DL0//EQ

A first order reaction has a rate constant

1.15 × 10–3 s–1. How long will 5 g of this reactant take to reduce to 3 g ?

//X

Integrated rate equation for first order reaction,
k = log
[R]0 = 5 g and [R] = 3 g, k = 1.15 × 10–3 s–1
Substituting above values in equation
t = log
t = log

//M2//QN6//SUB//DL0//EQ

Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, then calculate the rate constant of the reaction.

//X

Time required to decompose half of the reagent is 60 minutes, i.e. = 60 min.
For first order reaction, relation between half-life and rate constant k is
k = = 0.01155 min–1 OR
k = = 1.925 ×10– 4 s–1

//M2//QN7//SUB//DL0

Explain effect of temperature on rate constant. Or What will be the effect of temperature on rate constant?

//X

Most of the chemical reactions are accelerated by increase in temperature.
For example, in decomposition of N2O5, the time taken for half of the original amount of material to decompose is 12 min at 50°C, 5 h at 25°C and 10 days at 0°C.
In a mixture of potassium permanganate and oxalic acid, potassium permanganate gets decolourised faster at a higher temperature than that at a lower temperature.
It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.

//M3//QN8//SUB//DL0//EQ

The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K, calculate E a.

//X

log =
Here, T1 = 298K, T2 = 298 + 10K = 308K
R = 8.314 J K–1 mol–1
The rate of chemical reaction doubles for an increase of 10K in absolute temperature, therefore = 2.
Substituting the values in above equation, we get
log 2.0 =
Ea =
Ea =
Ea = 52897.77 J mol–1 = 52.89777 kJ mol–1

//M3//QN9//SUB//DL0//EQ

The activation energy for the reaction 2HI(g)→ H2 + I2(g) is 209.5 kJ mol–1 at

581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

//X

The fraction of molecules of reactants having energy equal to or greater than activation energy x =
Taking 10 base log on both the sides,
log x =
∴ log x =
∴ log x = –18.8323
∴ x = antilog (–18.8323) = .1677
∴ The fraction of molecules of reactants having energy equal to or greater than activation energy
x = 1.471 × 10–19