Ncert Solutions Chemistry Class 12th

4.17  (i) Average rate of reaction over interval is [change in concentration]/ [time taken] e.

[0.31 - 0.17] / [60-30] = 0.00467 mol L^-1 sec^-1

(ii) the pseudo first-order rate constant can be calculated by K = (2.303/t) log (C_i/C_t)

where K is Rate constant,

t is time taken,

C_i is initial concentration

C_t is Concentration at time t.

K = (2.303/30) log (0.55/0.31)

? K = 1.9 * 10^-2 sec^-1

(i) Average rate between 30 to 60 sec is 0.00467 mol L^-1sec^-1

(ii) Pseudo first order rate constant is 1.* 10^-2sec^-1

5.7

According to Hardy-Schulze law

'The greater the valence of the flocculating ion added, the greater is its power to cause precipitation.

As this law takes into consideration of only the charge present on the ion and not the size of the ion. So when the size of the atom is considered, smaller the size of an atom more will be its polarising power.

So Hardy-Schulze can be modified in terms of the polarising power of the flocculating ion as 'The greater the polarising power of the flocculating ion added, the greater is its power to cause precipitation.

4.16 Increase in temperature increases the rate constant of a reaction. as we know increase in temperature increases the rate of reaction to satisfy the equation Rate = k [concentration]ⁿ where n can be any real number. k have to increase as concentration is almost not changing over small temperature change.

Increasing the temperature by 10°c almost doubles the rate constant. This can be represented quantitatively by the help of the Arrhenius equation-

K = Ae^-Ea/RT, where k is rate constant, E_a is the activation energy, R is the universal gas constant, T is the absolute temperature.

5.6 

Desorption is a process in which substance (reactant + product) is released from the surface which is the opposite process of sorption.

The role of desorption in the process of catalysis is to make the surface of the solid catalyst-free for fresh adsorption of reactants on the solid surface for further reactions to take place.

The complex is an anion with chromium as central atom, 2 water molecules and 2 oxolate ions with -2 negative charge Balance overall charge as 0, we get oxidation state of Cr as:

X + 2 (0) + 2 (-2) = -1 X = + 3.

Name of compound: potassium diaquadioxolatochromate (III) trihydrate. Electronic configuration of Cr: 3d³, t_2g³

4.15 Let suppose reaction 

A? B; having rate law, Rate = K [A]²

(i) If the concentration of A is doubled then the rate will affect by the square of concentration i.e. rate will become 2² = 4 times.

(ii) If the concentration of A is halved, then the rate will affect the square of the concentration i.e (1/2)² = 1/4

(i) Rate becomes 4 times of initial Rate (ii) Rate becomes 1/4 times of initial Rate

4.14 Factors affecting rate of reaction-

1) Temperature (increasing temperature increases rate of reaction)

2) Concentration or pressure of reactants. (Increasing concentration or pressure increases rate of reaction)

3) Presence or absence of a (Adding catalyst mostly increases the rate of reaction)

4) The surface area of solid (If reaction is processing over solid reactant then increasing its surface area increases rate of reaction)

5) Nature of reactants

5.5

Ester hydrolysis is represented as:

Ester + Water → Acid + Alcohol

In this reaction the acid produced which is a product also acts as a catalyst and makes the reaction faster.

Such substances that act as catalysts in the same reaction in which they are obtained as products are known as Autocatalysts.

So, ester hydrolysis is slow in the beginning and becomes faster after some time as more acid is produced on the product side.

4.13 Rate of given chemical reaction will be represented as

-d [pCH₃OCH₃] / dt

Hence units of rate is bar min^-1

To find units of K, K = rate/ [pCH₃OCH₃]^3/2

The unit of k = bar ^-1/2min^-1.

Units of Rate- bar min^-1. and Units of Rate constant K : bar ^-1/2min ^-1