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(a) 10^-2 s^-1

(a) 1.44 x 10^-3 s^-1

For first order reaction,
Half life = \(\frac { 0.693 }{ k }\)
(i) t_1/2 = \(\frac { 0.693 }{ k }\) = 3.465 x 10^-3 s
(ii) t_1/2 = 0.3465 min
(iii) t_1/2 = \(\frac { 0.693 }{ 4 }\) = 0.1732 year

Option : (d) 0.4

On the basis of unit of rate constant (s^-1), the order of reaction is first order.

(c) SO₂Cl₂₍₁₎→ SO_2(g) + Cl_2(g)

(a) (i) only

(d) CH₃CHO_(g) → CH_4(g) + CO_(g)

1 is the order of isomerisation of cyclopropane to propene.

(b) Acid hydrolysis of an ester

(a) lodination of acetone in acid medium

(c) CH₃ CHO_(g) → CH_4(g) + CO_(g)

(c) (i) & (ii)

(d)  \(\frac{1}{16}\) g

in 140 days ⇒ initial concentration reduced to \(\frac{1}{2}\)g

in 280 days ⇒ initial concentration reduced to \(\frac{1}{4}\) g in 420 days ⇒ initial concentration reduced to \(\frac{1}{8}\)g

in 560 days ⇒ initial concentration reduced to \(\frac{1}{8}\)

The half life ola reaction is defined as the time required for the reactant concentration to reach one half its initial value

Let us consider the dimensation of a monomer M

2M → (M)₂

Rate = k [M]ⁿ 

Given that n =2 and [M] = 0.05 mol L^-1

Rate = 7.5 x 10^-3 mol L^-1 s^-1

Rate 7.5 x 10³ mol L^-1 s^-1

k = \(\frac{Rate}{[M]^n}\)

^{k = \(\frac{7.5\times10^{-3}}{(0.05)^2}\)} 

= 3 mol^-1 Ls^-1

(b) 2Na_(s)  + I_2(g) ⟶  2NaI_(s) 

Rate = k [x]^3/2[y]^1/2

overall order = \(\Big(\frac{3}{2}+\frac{1}{2}\Big)\) = 2

i.e., second order reaction. Since the rate expression does not contain the concentration of Z , the reaction is zero order with respect to Z.

(d) zero order

(c) Rate = k [A]. [B]

Order of the reaction = \(\frac{1}{2}\) + 2 = 2 \(\frac{1}{2}\) or 0.5

The rate of a reaction is affected by the following factors. 

1. Nature and state of the reactant 

(a) A chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product. The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants. 

(b) Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants. For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine. 

2. Concentration of the reactant 

The rate of the reaction increases with the increase in the concentration of the reactants. According to collision theory, the rate of the reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate. 

3. Effect of surface area of the reactant: 

In heterogeneous reactions, the surface areas of the solid reactants plays an important role in deciding the rate. For a given mass of a reactant, when the particle size decreases surface area increases.

Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of reaction is increased. For example, powdered calcium carbonate reacts much faster with dilute HCI than with the same mass of CaCOl as marble

4. Temperature: 

For many reactions near room temperature, the reaction rate tends to double when the temperature is increased by 10°C . For eg, Reaction between H₂ and O₂ to form H₂O take place only when an electric spark is passed. So when the temperature increases, the rate of the reaction also increases. 

5. Effect of presence of catalyst 

(a) A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change. They may participate in the reaction, but again regenerated and the end of the reaction. 

(b) In the presence of a catalyst, the energy of activation is lowered and hence greater number of molecules can cross the energy barrier and change over to products,thereby increasing the rate of the reaction.

(b) 2.5 hours

(a) first order

1. Paracetamol is a well known antipyretic and analgesic that is prescribed in cases of fever and body pain.

2. Paracetamol has a half life of 2.5 hours within the body. (Le) the plasma concentration of the drug is halved after 2.5 hours. So after 10 hours (4 half lives), only 6.25% of drug remains. Based on this, the dosage and frequency will be decided.

3. In the case of paracetamol, it is usually prescribed to take once in 6 hours.

The rate of the reaction is affected by the following factors. 

1. Nature and state of the reactant 

2. Concentration of the reactant 

3. Surface area of the reactant 

4. Temperature of the reaction 

5. Presence of a catalyst

1. In heterogeneous reactions, the surface area of the solid reactants play an important role in deciding the rate. 

2. For a given mass of a reactant, when the particle size decreases surface area increases. Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of reaction is increased.

3. For example, powdered calcium carbonate reacts much faster with dilute HCl than with the same mass of CaCO₃ as marble.

Rate of reaction increases with increase of temperature.

Rate constant increases nearly by two times.

The time in which the conc.of a reactant is reduced to one half of its initial conc. is called half life of a reaction (t₁/2). 

Option : (c) atm s^-1

Option : (c) \(-\frac{1}{3}\)\(\frac{d[A]}{dt}\)

Option : (c) twice the rate of consumption of A.

Concentration, temperature, presence of catalyst and light, surface area of reactants.

Show that rate of a reaction can be expressed in terms of each reactant and product- 

Rate of disappearance of reactant = rate of reaction × stoichiometric coefficient of the reactant R

ate of appearance of product = rate of reaction × stoichiometric coefficient of the product

\(\frac{-d[N_2]}{dt} = \frac{-1}{3} \frac{d[H_2]}{dt} = \frac{+1}{2} \frac{d[NH_3]}{dt}\)

Option : (c) 2B+C → 3A

The units of rate constant (k) for the first order reaction is per time (or s^-1).

Option : (b) mol dm^-3t^-1

Option : (a) zero.

Option : (a) increases

(i) Pseudo first order reaction: If a reaction is not truly of the first order but under certain conditions become reaction of first order is called pseudo fust order reaction.

(ii) Half-life period of a reaction (t_1/2): Half-life of a reaction is the time in Which the concentration of a reactant is reduced to half of its original value.

Option : (b) k = \(-\frac{1}{t}ln\frac{[A]_t}{[A]_0}\)

Given : 

Rate constant of the reaction = k = 3.74 x 10^-2 M^-2 s^-1

[A] =0.108 M, 

[B] = 0.132M, 

[C] = 0.124 M 

Rate of the reaction = R = ?

By rate law,

R = k [A]² x [B] 

= (0.108)² x 0.132 

= 1.54 x 10 Ms^-1

(Concentration of C need not be considered since it is a product.)

∴ Rate of reaction = 1.54 x 10^-3 Ms^-1

(a) For the reaction,

2H₂O₂ → 2H₂O + O₂

Since the rate law expression given is,

Rate = k [H₂O₂]

Hence,

The reaction is of first order.

(b) For the reaction,

NO₂ + CO → NO + CO₂

Since the rate law given is Rate = k [NO₂]²,

The reaction is second order with respect to NO₂ and zero order with respect to CO.

Hence the net order of the reaction is,

n = nNO₂ + nco = 2 + 0 = 2

(c) For the reaction,

2NO + O₂ → 2NO₂

Since the rate law expression given is, 

Rate = k[NO]² x [O₂]

The reaction is second order with respect to NO and first order with respect to O₂.

Hence,

The overall order of reaction is n = nNO₂ + no₂ = 2 + 1 = 3.

(d) For the reaction, 

CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g)

By rate law,

Rate = k [CHCl₃] [Cl₂]

Reaction is first order with respect to CHCl₃ and first order with respect to Cl₂. 

Hence,

The overall order is, 

n = ncHcl₃ + ncl₂ 

= 1 + 1 = 2.

(i) 2 

(ii) 2 

(iii) 3/2 

(iv) 1

Rate = k x [A]² x [B] 

= 6.25 x 1² x 0.2

Rate = 1.25 x 10² mol dm^-3 s^-1

(a) For first order reaction, half-life period t_1/2 is, t_1/2 \(=\frac{0.693}{k}\) where k is the rate constant.

(b) For zeroth-order reaction, half half period (t_1/2) is, t_1/2 = \(\frac{[A]_0}{2k}\) where k is the rate constant and [A]₀ is initial concentration of the reactant.

Given :

2N₂O_5(g) → 4NO_2(g) + O_2(g)

(a) Rate of formation of NO₂ at time,

t = \(\frac{d[NO_2]}{dt}\)

Rate of formation of O₂ at time,

t = \(\frac{d[O_2]}{dt}\)

They are related to each other through rate of reaction.

∴ Rate of reaction = \(\frac{1}{4}\)\(\frac{d[NO_2]}{dt}\)\(\frac{d[O_2]}{dt}\)

(b) Rate of consumption of N₂O₅ at time t,

= \(-\frac{d[N_2O_5]}{dt}\)

Rate of reaction,

= \(-\frac{1}{2}\)\(\frac{d[N_2O_5]}{dt}\) 

= \(\frac{d[O_2]}{dt}\)

In general,

Rate of reaction,

= \(-\frac{1}{2}\)\(\frac{d[N_2O_5]}{dt}\)

= \(\frac{1}{4}\)\(\frac{d[NO_2]}{dt}\) 

= \(\frac{d[O_2]}{dt}\)