Q – 4.1. 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.

Q – 4.2. In a reaction, 2 A ⟶ Products, the concentration of A decrases from 0.5 mol L-1 to 0.4 mol L -1 in 10 minutes. Calculate the rate during this interval.

Q – 4.3. For a reaction, A + B ⟶ Product , the rate law is given by :

r = K [A]1/2 [B]2 what is the order of the reaction?

r = K [A]1/2 [B]2 what is the order of the reaction?

Q – 4.4. the conversion of the molecules X to Y follows second kinetics. If the concentration X is increased to three times, how will it affect the rate the of formation of Y?

Ans. For the reaction, X ⟶ Y, as it follows second order kinetics, the rate law equation will be

Rate = k [X]2 = ka2 {if [X] = a mol L-1}

If concentration of X is increased three times, now, [X] = 3 a mol L -1

؞ Rate k (3 a)2 = 9 ka2

Thus, the rate of reaction will become 9 times. Hence, the rate of formation of Y will increase 9 times.

Rate = k [X]2 = ka2 {if [X] = a mol L-1}

If concentration of X is increased three times, now, [X] = 3 a mol L -1

؞ Rate k (3 a)2 = 9 ka2

Thus, the rate of reaction will become 9 times. Hence, the rate of formation of Y will increase 9 times.

Q – 4.5. A first order reaction has a rate constant 1.15 × 10-3s-1. How long will 5 g of this reactant take to reduce to 3 g?

Ans. Here, [A]0 = 5 g, [A] = 3 g, k = 1.15 × 10-3 s-1. As the reaction is of list of 1st order,

= 2.0 × 103 × 0.2219 s = 443.8 s = 444 s.

= 2.0 × 103 × 0.2219 s = 443.8 s = 444 s.

Q – 4.6. Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

Q – 4.7. What will be the effect of temperature on rate constant?

Ans. Rate constant of a reaction is nearly doubled with rise in temperature by 100. The exact dependence of the rate constant on temperature is given by Arrhenius equation, k = A e^(〖-E〗_a/RT) where A is called frequency factor and E0 is the activation energy of the reaction.

Q – 4.8. the rate of the particular reaction doubles when temperature changes from 270 C to 370 C. calculate the energy of activation of such a reaction.

Ans. here, we are given that

when T1 = 270C = 300 K,

K1 = k (say). When T2 = 370C = 310 K, k2 = 2k

Substituting these values in the equation:

This on solving gives Ea = 53598.6 J mol-1 = 53.6 kJ mol-1

when T1 = 270C = 300 K,

K1 = k (say). When T2 = 370C = 310 K, k2 = 2k

Substituting these values in the equation:

This on solving gives Ea = 53598.6 J mol-1 = 53.6 kJ mol-1

Q – 4.9. The activation energy for the reaction, 2 HI (g) ⟶ H2 (g) + I2 (g), is 209.5 kJ mol-1 at 581 K.

Calculate the fraction of molecules of reactants having energy equal to or greater then activation energy.

Calculate the fraction of molecules of reactants having energy equal to or greater then activation energy.