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Page No 116: - Chapter 4 Chemical Kinetics Intext Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024]


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,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6323/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_m70f0906.gif

Where,

A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

Ea 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 Ea.

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:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6325/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_75a3d42e.gif

We get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6325/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_41219bce.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6325/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_782090af.gif

= 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:

Ea = 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

Also Read : Page-No-146:-Chapter-5-Surface-Chemistry-Exercise-Solutions-class-12-ncert-solutions-Chemistry

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