Question 1

For the reaction 2 SO _2+ O _2 2 SO _3, the rate of disappearance of O _2 is 2 10^ -4 ~mol ~L ^ -1 s ^ -1 . The rate appearance of SO _3 is

Accepted Answer

2 SO _2+ O _2 2 SO _3 Rate of reaction =- 1 2 d [ SO _2 ] d t =- d [ O _2 ] dt = 1 2 d [ SO _3 ] d t Rate of appearance of SO _3, aligned d [ SO _3 ] d t & =2 [ -d [ O _2 ] d t ]=2 (2 10^ -4 ) \ & =4 10^ -4 ~mol ~L ^ -1 ~s ^ -1 aligned

Question 2

If for a first order reaction, the value of A and E_a are 4 10^ 13 ~s ^ -1 and 98.6 ~kJ ~mol ^ -1 respectively, then at what temperature will its half life period be 10 minutes?

Accepted Answer

According to Arrhenius equation, k= A- E_a 2.303 R T ...(i) For 1^ st order reaction, k= 0.693 t_ 1 / 2 = 0.693 600 s =1.1 10^ -3 ~s ^ -1 Substitute the value of k in Eq. (i) log (1.1 10^ -3 )= (4 10^ 13 )- 98.6 10^3 2.303 8.314 T Solving for T, T=311.15 ~K

Question 3

In the chemical reaction A B, what is the order of the reaction? Given that, the rate of reaction doubles if the concentration of A is increased four times.

Accepted Answer

We know, r_1=k[A]^ n ....(i) If conc. of A is increased four times rate of reaction on doubles. 2 r_1=k[4 A]^ n ....(ii) Divide Eq. (ii) by (i). 2 r_i r_1 = k(4 A)^n k(A)^n 2=(4)^ n 2 n =(2)^ 2 n n=0.5

Question 4

Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from 300 K to 600 K .

Accepted Answer

( k_2 k_1 )= E_a 2.303 R [ 1 T_1 - 1 T_2 ] or, ( 2 k k )= E_a 2.303 8.3 [600-300] 300 600 E_a 3.45 ~kJ / mol

Question 5

In the reaction, A products, If the concentration of the reactant is doubled rate of the reaction remains unchanged. The order of the reaction with respect to A is

Accepted Answer

r_1=k[A]^n and r_2=k[2 A]^n r _1= r _2 k [ A ]^ n = k (2[ ~A ])^ n or, 2^n=1 or, 2^ n =2^ n =0 It is a zero order reaction.

Question 6

In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is

Accepted Answer

As the concentration of reactant decreases from 0.8 to 0.4 in 15 minutes, hence the t_ 1 / 2 is 15 minutes. To fall the concentration from 0.1 to 0.025 , we need two half lives, i.e., 30 minutes.

Question 7

For the decomposition of azomethane, CH _3 ~N _2 CH _3( ~g ) CH _3 CH _3( ~g )+ N _2( ~g ) a first order reaction, the variation in partial pressure with time at 600 ~K is given as The half life of th

Accepted Answer

For first order reaction aligned & k= 1 t ( P_0 P ) In ( P_0 P )=k t \ & ( P P_0 )=-k t aligned On comaring with straight line equation y = mx aligned & k = slope =3.465 10^4 \ & t _ 1 / 2 = 2 k = 0.693 3.465 10^4 =2 10^ -5 ~s aligned

Question 8

Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is

Accepted Answer

As per Arrhenius equation (k=A e ^ -E_a / R T ), the rate constant increases exponentially with temperature.

Question 9

In a 1 st order reaction, reactant concentration C varies with time t as :

Accepted Answer

Answer option index: 1

Question 10

XmL of H _ 2 gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is :

Accepted Answer

Under identical conditions, r_ 1 r_ 2 = M _ 2 M _ 1 As rate of diffusion is also inversely proportional to time, we will have, t_ 2 t_ 1 = M _ 2 M _ 1 (A) Thus, For He , t_ 2 = 4 2 (5 s)=5 2 s 10 s; (B) For O _ 2 , t_ 2 = 32 2 (5 s)=20 s (C) For CO , t_ 2 = 28 2 (5 s) 25 s; (D) For CO _ 2 , t_ 2 = 44 2 (5 s) 55 s #

Question 11

A radioactive isotope having a half - life period of 3 days was received after 12 days. If 3 ~g of the isotope is left in the container, what would be the initial mass of the isotope?

Accepted Answer

Given t_ 1 / 2 =3 Total time T =12 No. of half lives (n)= 12 3 =4 ( 1 2 )^ n = N N _ o ( 1 2 )^ 4 = 3 ~N 3 ~N = 1 16 N=48

Chemical Kinetics — Previous Year Questions & Solutions