Page No 116: - Chapter 4 - Chemical Kinetics - Intext Solutions - Ncert Solutions class 12 - Chemistry

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Question 4.7:

What will be the effect of temperature on rate constant?

Answer:

The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,

Where,

A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

E_a is the activation energy

Question 4.8:

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_a.

Answer:

It is given that T₁ = 298 K

∴T₂ = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k₁ = k and that of k₂ = 2k

Also, R = 8.314 J K⁻¹ mol⁻¹

Now, substituting these values in the equation:

We get:

= 52897.78 J mol⁻¹

= 52.9 kJ mol⁻¹

Note: There is a slight variation in this Answer and the one given in the NCERT textbook.

Question 4.9:

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?

Answer:

In the given case:

E_a = 209.5 kJ mol⁻¹ = 209500 J mol⁻¹

T = 581 K

R = 8.314 JK⁻¹ mol⁻¹

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

x=e-Ea/RT⇒In x=-EaRT⇒log x=-Ea2.303RT⇒log x=-209500 J mol-12.303×8.314 JK-1mol-1×581=-18.8323Now, x =Antilog -18.8323             

 = 1.471×10-19