Q: If the rate constant of a reaction is 0.03 ~s ^ -1 , how much time does it take for 7.2 ~mol ~L ^ -1 concentration of the reactant to get reduced to 0.9 ~mol ~L ^ -1 ? (Given : 2=0.301 )

aligned & K =0.03 ~s ^ -1 (First order reaction) \\ & A _0=7.2 ~mol ~L ^ -1 \\ & ~A _ t =0.9 ~mol ~L ^ -1 \\ & 7.2 t _ 1 / 2 3.6 t _ 1 / 2 1.8 t _ 1 / 2 0.9 \\ & t =3 t _ 1 / 2 = 3 0.693 0.03 =69.3 ~s aligned

Question 1

2 A k B is a zero-order reaction, where k=1.0 mol L^ -1 min^ -1 . If the initial concentration of A is 2 M, then the time taken to complete 75\% of the reaction will be

Accepted Answer

For a zero-order reaction 2 A B , the rate of reaction is expressed as: Rate = - 1 2 d[ A ] dt = k - d[ A ] dt = 2k Integrating this equation, we get the integrated rate law: [ A ]_0 - [ A ]_t = 2kt Given: Initial concentration, [ A ]_0 = 2 M Rate constant, k = 1.0 mol L^ -1 min^ -1 For 75\% completion of the reaction, the amount of A reacted is: [ A ]_0 - [ A ]_t = 75 100 2 = 1.5 M Substituting these values into the integrated rate equation: 1.5 = 2 1.0 t 2t = 1.5 t = 0.75 min

Question 2

Match List I with List II : List I (Order of reaction) List II (Unit of rate constant) A. Zero order I. mol ^ -1 \ L \ s ^ -1 B. First order II. mol ^ -2 \ L ^ 2 \ s ^ -1 C. Second order III. s ^ -1 D

Accepted Answer

The unit of rate constant k for an n-th order reaction is given by ( mol \ L ^ -1 )^ 1-n \ s ^ -1 . For zero order reaction (n=0): Unit = ( mol \ L ^ -1 )^ 1-0 \ s ^ -1 = mol \ L ^ -1 \ s ^ -1 (Matches A with IV) For first order reaction (n=1): Unit = ( mol \ L ^ -1 )^ 1-1 \ s ^ -1 = s ^ -1 (Matches B with III) For second order reaction (n=2): Unit = ( mol \ L ^ -1 )^ 1-2 \ s ^ -1 = mol ^ -1 \ L \ s ^ -1 (Matches C with I) For third order reaction (n=3): Unit = ( mol \ L ^ -1 )^ 1-3 \ s ^ -1 = m

Question 3

For an elementary chemical reaction, the Arrhenius plot is given below. If the energy of activation is 6.64 kJ mol^ -1 and R =8.3 J K^ -1 mol^ -1 , the temperature at which the rate constant becomes e

Accepted Answer

According to the Arrhenius equation: k = A - E_a RT For a plot of k versus 1/T, the y-intercept corresponds to A. From the given graph, the line intersects the y-axis at 6. Therefore, A = 6. We are given: E_a = 6.64 kJ mol ^ -1 = 6640 J mol ^ -1 R = 8.3 J K ^ -1 mol ^ -1 k = e^2 min ^ -1 , which gives k = (e^2) = 2 Substituting these values into the Arrhenius equation: 2 = 6 - 6640 8.3 T 6640 8.3 T = 6 - 2 6640 8.3 T = 4 T = 6640 4 8.3 T = 6640 33.2 T = 200 K Answer: 200 K

Question 4

Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, T (K). k = 14.34 - 1.25 10^ 4 T The energy of activation in kcal mol^ -1 for the reacti

Accepted Answer

The Arrhenius equation is given by: k = A - E_a RT The given equation is: k = 14.34 - 1.25 10^ 4 T Comparing the two equations: E_a R = 1.25 10^ 4 E_a = 1.25 10^ 4 R Substituting R = 1.987 cal mol^ -1 K^ -1 : E_a = 1.25 10^ 4 1.987 E_a = 24837.5 cal mol^ -1 E_a = 24.8375 kcal mol^ -1 24.84 kcal mol^ -1 Answer: 24.84

Question 5

For a certain reaction R Product, the plot of concentration [R] vs time has a negative slope as shown. The order of reaction is :

Accepted Answer

From the given graph, the plot of concentration [R] versus time t is a straight line with a negative slope. The equation for this straight line can be written as: y = mx + c [R] = ( slope ) t + intercept Given that the intercept is [R_0] and the slope is -k, the equation becomes: [R] = -kt + [R_0] This equation represents the integrated rate law for a zero-order reaction. For a zero-order reaction, the rate of reaction is independent of the concentration of the reactant: Rate = - d[R] dt = k[R]^

Question 6

If the half-life ( t _ 1 / 2 ) for a first order reaction is 1 minutes, then the time required for 99.9 \% completion of the reaction is closest to:

Accepted Answer

For 1^ st order reactions : aligned t _ 99.90 \% & =10 t _ 1 / 2 \ t _ 99.9 \% & =10 1 \ & =10 minute aligned

Question 7

If the rate constant of a reaction is 0.03 ~s ^ -1 , how much time does it take for 7.2 ~mol ~L ^ -1 concentration of the reactant to get reduced to 0.9 ~mol ~L ^ -1 ? (Given : 2=0.301 )

Accepted Answer

aligned & K =0.03 ~s ^ -1 (First order reaction) \ & A _0=7.2 ~mol ~L ^ -1 \ & ~A _ t =0.9 ~mol ~L ^ -1 \ & 7.2 t _ 1 / 2 3.6 t _ 1 / 2 1.8 t _ 1 / 2 0.9 \ & t =3 t _ 1 / 2 = 3 0.693 0.03 =69.3 ~s aligned

Question 8

Following data is for a reaction between reactants A and B : The order of the reaction with respect to A and B, respectively, are

Accepted Answer

Let the rate equation is Rate = k [ A ]^ x [ B ]^ y Therefore, we can write 2 10^ -3 =k[0.1] [0.1 ^y ....(i) 4 10^ -3 =k[0.2]^x[0.1]^y...(ii) 1.6 10^ -2 = k [0.2]^ [0.2]^ y ...(iii) (ii) (i); 4 10^ -3 2 10^ -3 = k[0.2]^x[0.1]^y k[0.1]^x[0.1]^y 2 1 = (0.2)^x (0.1)^x = ( 2 1 )^x x=1 (ii) (iii); 4 10^ -3 1.6 10^ -2 = k[0.2]^ x [0.1]^y k [0.2]^ x [0.2]^ y 1 4 = (0.1)^y (0.2)^y = ( 1 2 )^y y=2 Rate =k[A]^1[B]^2 First order with respect to A while second order with respect to B .

Question 9

Which of the following plot represents the variation of k versus 1 T in accordance with Arrhenius equation?

Accepted Answer

Using Arrhenius equation, k = Ae ^ - E _ a / R T k = Ae ^ - E _ a / R T k= A- E_a R T y=c+m x, slope (m)=- E_a R Intercept = A

Question 10

Activation energy of any chemical reaction can be calculated if one knows the value of

Accepted Answer

To calculate value of E_a Equation used is ( k_2 k_1 )= E_a 2.303 R ( 1 T_1 - 1 T_2 ) Hence E_a can be calculated if value of rate constant k is known at two different temperatures T_1 and T _2.

Question 11

Rate constants of a reaction at 500 K and 700 K are 0.04 ~s ^ -1 and 0.14 ~s ^ -1 , respectively; then, activation energy of the reaction is : (Given: 3.5=0.5441, R =8.31 ~J ~K ^ -1 ~mol ^ -1

Accepted Answer

K =A e^ -E_a / R T After taking In both side K= A- E_a R T In K _1= A - E _ a RT _1 at temp. T _1...(i) In K _2= A - E _ a RT _2 at temp. T _2,,(ii) (ii) - (i) In K_2- InK K_1= E_a R [ 1 T_1 - 1 T_2 ] K _2 ~K _1 = E _ a R [ 1 500 - 1 700 ] 0.14 0.04 = E_a R [ 700-500 500 700 ] 14 4 = E_a R [ 200 500 700 ] 3.5= E_a 2.303 R [ 1 250 7 ] 0.5441= E_a 2.303 8.31 [ 1 250 7 ] E_a=0.5441 8.31 250 7 2.303 =0.5441 83.1 25 7 2.303 =18222.65 18219 ~J

Question 12

The rate of a reaction quadruples when temperature changes from 27^ C to 57^ C . Calculate the energy o activation. Given R =8.314 ~J ~K ^ -1 ~mol ^ -1 , 4=0.6021

Accepted Answer

aligned & ( k _2 k _1 )= E _ a 2.303 R ( 1 ~T _1 - 1 ~T _2 ) \ & ( 4 1 )= E _ a 2.303 R ( 1 300 - 1 330 ) \ & E _ a = ( (4)) 2.303 8.314 300 330 30 \ & =3.804 10^4 ~J / mol \ & =38.04 ~kJ / mol aligned

Question 13

Which plot of k vs 1 ~T is consistent with Arrhenius equation?

Accepted Answer

The Arrhenius equation is given as aligned & k=A e^ - E_a R T \ & k= A- E_a R T aligned In k v / s 1 ~T gives a straight line graph with slope =- E_ a R and intercept = A

Chemical Kinetics — Previous Year Questions & Solutions