# NCERT 12 Chemistry Chemical Kinetics Chapter 4 exercises 1

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? 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.
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.
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
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. Updated: December 4, 2021 — 3:11 pm