Q: Consider a reaction A+R Product. The rate of this reaction is measured to be k[A][R]. At the start of the reaction, the concentration of R,[R]_0, is 10 -times the concentration of A,[A]_0. The reactio

aligned & A + R Product \\ & t =0 ~A _0 10 ~A _0 \\ & t = t 0.6 ~A _0 9.6 ~A _0 \\ & Rate = k [ A ][ R ] \\ & Rate _1= k (0.6 ~A _0 ) 9.6 ~A _0 \\ & ~A + R Product \\ & t =0 ~A _0 10 ~A _0( excess ) \\ & t = t 0.6 ~A _0 10 ~A _0 \\ & Rate = k ^ [ A ], k ^ = k [ R ] \\ & Rate _2= ( k 10 ~A _0 ) (0.6 ~A _0 ) \\ & 100 Rate Rate _1 = (0.6 10-0.6 9.6 0.6 9.6 100=4.1666 aligned

Q: The plot of logk f versus 1 T for a reversible reaction A ( g ) ⇌ P ( g ) is shown. Pre-exponential factors for the forward and backward reactions are 10 15 s - 1 and 10 11 s - 1 ,

From the question A f = 10 15 , A b = 10 11 , f = Forward reaction logK at 500 K = 6 b = Backward reaction logk f at 500 K = 9 (from graph) logk b at 500 K : logK = log k f k b since ⇒ K = k f k b 6 = logk f - logk b 6 = 9 - logk b log ⁡ k b = 3 at 500 K log ⁡ k 2 k 1 = − E a R 1 T 2 − 1 T 1 k b = A b e − E ab RT ln ⁡ k b = ln ⁡ A b e &#8722

Q: Match the rate expressions in LIST-I for the decomposition of $ X $ with the corresponding profiles provided in LIST-II. $X_s$ and $ k $ are constants having appropriate units.

I rate = k X X S + X Case-1: X > > X s ; X + X s ≈ X rate = k X X = k (Zero order w.r.t. X ) I → P , S Case-2: X ≪ X s ; X + X s ≈ X s ∴ rate = k X X S = k ' X ( 1 st order w.r.t. X ) ∴ I → Q , T Case-3: X ≈ X S rate = k X X s + X In this case curve- R given in List-II will match. ∴ I → P , Q , R , S , T &

Q: For the following reaction 2 X + Y → k P the rate of reaction is d [ P ] dt = k [ X ] . Two moles of X are mixed with one mole of Y to make 1 . 0 L of solution. At 50 s

$ array lllll & 2 X + & y & & P \\ t =0 & 2 & 1 & & 0 \\ t =50 & 2-2 0.5 & 1-0.5 & & \\ & 1mole & 0.5 & & array $ rate $=- 1 2 d x d t =- d y d t = d P d t =K[X]$ $- 1 2 d x d t =K[X]$ $- d x d t =2 K[X]=K^1[X]$ Half life is $t=50 sec $ $2 K= 0.653 L 50 $ $K= 0.6932 100 =6.332 10^ -3 $ $t=50 sec $ $- d x d t =2 K[X]$ $- d x d t =2 6.332 10^ -3 1$ $=13.864 10^ -3 ~mole / L / Sec $ $- d y d t =K[X]=6.332 10^ -3 ( 1 2 )$ $=3.46 10^ -3 ~mole / L / Sec ^ -1 $

Q: U 92 238 is known to undergo radioactive decay to form Pb 82 206 by emitting alpha and bet a particles. A rock initially contained 68 × 10 - 6 g of U 92 238 . If the number of alpha particles tha

92 238 U → 82 206 Pb + 2 4 He + 6 − 1 0 β 0 . 286 × 10 - 6 mole 0 . 286 × 10 - 6 × N a = 0 . 286 × 6 . 02 × 10 17 = 1 . 72 × 10 17 After 3 half lives N = No ( 2 ) 3 = 1 . 72 × 10 17 8 = 0 . 215 × 10 17 So, no. of molecule of uranium decayed = [ 1 . 72 - 0 . 215 ] × 10 17 = 1 . 5 × 10 17 - So No. of α particle produced = 8 × 1 . 5 × 10 17 = 12 × 10 17 = 1 . 2 × 10 18 &nb

Question 1

For a reversible reaction R P , at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constants k_f and k_b, respectively. At time zero, t

Accepted Answer

For the reversible first-order reaction R P , at equilibrium, the rate of the forward reaction equals the rate of the backward reaction. k_f [ R ]_ eq = k_b [ P ]_ eq Given k_b = 4k_f, we can substitute this into the equilibrium condition: k_f [ R ]_ eq = 4k_f [ P ]_ eq [ R ]_ eq = 4[ P ]_ eq From the conservation of mass, the total concentration remains constant at any time t: [ R ]_ eq + [ P ]_ eq = [ R ]_0 Substituting [ R ]_ eq = 4[ P ]_ eq into the mass balance equation gives: 4[ P ]_ eq +

Question 2

For a first-order reaction R P at a given temperature, k is the rate constant. For this reaction, at the given temperature, the concentrations of R and P at a time t are [ R ] and [ P ], respectively.

Accepted Answer

For a first-order reaction R P , the rate law is given by: - d [ R ] d t = d [ P ] d t = k[ R ] Integrating this, the concentration of the reactant at time t is [ R ] = [ R ]_0 e^ -kt . The concentration of the product at time t is [ P ] = [ R ]_0 - [ R ] = [ R ]_0 (1 - e^ -kt ). The plot of [ P ] versus t follows the equation [ P ] = [ R ]_0 (1 - e^ -kt ). This represents a curve that starts from the origin and is concave down, approaching a constant value [ R ]_0. The graph in (A) is concave u

Question 3

Consider a reaction A+R Product. The rate of this reaction is measured to be k[A][R]. At the start of the reaction, the concentration of R,[R]_0, is 10 -times the concentration of A,[A]_0. The reactio

Accepted Answer

aligned & A + R Product \ & t =0 ~A _0 10 ~A _0 \ & t = t 0.6 ~A _0 9.6 ~A _0 \ & Rate = k [ A ][ R ] \ & Rate _1= k (0.6 ~A _0 ) 9.6 ~A _0 \ & ~A + R Product \ & t =0 ~A _0 10 ~A _0( excess ) \ & t = t 0.6 ~A _0 10 ~A _0 \ & Rate = k ^ [ A ], k ^ = k [ R ] \ & Rate _2= ( k 10 ~A _0 ) (0.6 ~A _0 ) \ & 100 Rate Rate _1 = (0.6 10-0.6 9.6 0.6 9.6 100=4.1666 aligned

Question 4

A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into Pb -206 while the rest of it remains undisintegrated. When the age of the sample is P

Accepted Answer

_ 92 U ^ 238 _ 82 ~Pb ^ 206 +8 _2 He ^4+6_ -1 ^0 t =0 ~N _0= ( 1 238 + 7 206 ) moles t = tN _ t = 1 238 moles x = 7 206 moles Life of sample $ t$ years $[A]_0 $ Initial mole of $ U -238$ $[A]]_t $ Final mole of U-238 $ aligned & |A|_0 |A|_t = 1 228 + 7 206 1 238 \ & = 0.0042+0.0340 0.0042 \ & =9.1 \ & = 2.303 2 t 4.5 10^9 =2.303 9.1 \ & t=14.27 10^9 years \ & =142.7 10^8 years \ & P=142.7 \ & P=143 aligned $

Question 5

Consider the following reaction, 2 H _2( ~g )+2 NO ( g ) N _2( ~g )+2 H _2 O ( g ) which follows the mechanism given below: 2 NO ( g ) k_1 k_ -1 N _2 O _2( ~g ) (fast equlibrium) N _2 O _2( ~g )+ H _2

Accepted Answer

Rate law = k _2 [ ~N _2 O _2 ] [ H _2 ] [ slowest step of reaction is RDS ] k _1 k _ -1 = [ N _2 O _2 ] [ NO ]^2 [ N _2 O _2 ]= k _1 k _ -1 [ NO ]^2 Rate = k _2 k _1 k _ -1 [ NO ]^2 [ H _2 ] Order of reaction is (3)

Question 6

The plot of logk f versus 1 T for a reversible reaction A ( g ) &#8652; P ( g ) is shown. Pre-exponential factors for the forward and backward reactions are 10 15 &#160; s - 1 and 10 11 &#160; s - 1 ,

Accepted Answer

From the question A f = 10 15 , A b = 10 11 , f = &#160;Forward reaction&#160; logK &#160;at&#160; 500 K = 6 b = &#160;Backward reaction&#160; logk f &#160;at&#160; 500 K = 9 &#160;(from graph)&#160; logk b &#160;at&#160; 500 K : logK = log k f k b &#160;&#160; &#160;since&#160; &#8658; K = k f k b 6 = logk f - logk b 6 = 9 - logk b log &#8289; k b = 3 &#160;at&#160; 500 K log &#8289; k 2 k 1 = &#8722; E a R 1 T 2 &#8722; 1 T 1 k b = A b e &#8722; E ab RT ln &#8289; k b = ln &#8289; A b e &#8722

Question 7

Match the rate expressions in LIST-I for the decomposition of $ X $ with the corresponding profiles provided in LIST-II. $X_s$ and $ k $ are constants having appropriate units.

Accepted Answer

I  rate  = k X X S + X Case-1:  X &#62; &#62; X s ; X + X s &#8776; X rate  = k X X = k     (Zero order w.r.t. X ) I &#8594; P , S   Case-2:  X &#8810; X s ; X + X s &#8776; X s &#8756;  rate  = k X X S = k ' X    ( 1 st  order w.r.t. X ) &#8756; &#160; &#160; I &#8594; Q , T Case-3:  X &#8776; X S rate  = k X X s + X In this case curve- R  given in List-II will match. &#8756; &#160; I &#8594; P , Q , R , S , T &

Question 8

For the following reaction  2 X + Y &#8594; k P  the rate of reaction is d [ P ] dt = k [ X ] . Two moles of X are mixed with one mole of Y to make 1 . 0 &#160; L of solution. At 50 &#160; s

Accepted Answer

$ array lllll & 2 X + & y & & P \ t =0 & 2 & 1 & & 0 \ t =50 & 2-2 0.5 & 1-0.5 & & \ & 1mole & 0.5 & & array $ rate $=- 1 2 d x d t =- d y d t = d P d t =K[X]$ $- 1 2 d x d t =K[X]$ $- d x d t =2 K[X]=K^1[X]$ Half life is $t=50 sec $ $2 K= 0.653 L 50 $ $K= 0.6932 100 =6.332 10^ -3 $ $t=50 sec $ $- d x d t =2 K[X]$ $- d x d t =2 6.332 10^ -3 1$ $=13.864 10^ -3 ~mole / L / Sec $ $- d y d t =K[X]=6.332 10^ -3 ( 1 2 )$ $=3.46 10^ -3 ~mole / L / Sec ^ -1 $

Question 9

Which of the following plots is(are) correct for the given reaction? ( [ P ] 0 is the initial concentration of P )

Accepted Answer

this is 1 st order reaction For 1 st T 1 / 2 = 0 . 693 k So independent of initial concentration C = C 0 e - kt ⇒ tn C C 0 = - k ⇒ ln P [ P ] 0 = - k So graph D is incorrect.

Question 10

U 92 238 is known to undergo radioactive decay to form Pb 82 206 by emitting alpha and bet a particles. A rock initially contained 68 &#215; 10 - 6 g of U 92 238 . If the number of alpha particles tha

Accepted Answer

92 238 U &#8594; 82 206 Pb + 2 4 He + 6 &#8201; &#8722; 1 &#160; &#160; &#160; &#160; 0 &#946; 0 . 286 &#215; 10 - 6 mole 0 . 286 &#215; 10 - 6 &#215; N a = 0 . 286 &#215; 6 . 02 &#215; 10 17 = 1 . 72 &#215; 10 17 After 3 half lives N = No ( 2 ) 3 = 1 . 72 &#215; 10 17 8 = 0 . 215 &#215; 10 17 So, no. of molecule of uranium decayed = [ 1 . 72 - 0 . 215 ] &#215; 10 17 = 1 . 5 &#215; 10 17 - So No. of &#945; particle produced = 8 &#215; 1 . 5 &#215; 10 17 = 12 &#215; 10 17 = 1 . 2 &#215; 10 18 &nb

Question 11

The decomposition reaction 2 N 2 O 5 g &#8594; &#160; &#8710; &#160; 2 N 2 O 4 g + O 2 g is started in a closed cylinder under the isothermal isochoric condition at an initial pressure of 1 &#160; atm

Accepted Answer

P 0 = 1 atm , P t = 1 .45 atm , P ∝ = 1 .5 atm From the unit of K = 5 × 10 - 4 s - 1 , The reaction is first-order t = 2.303 2 × 5 × 10 - 1 log ⁡ 1.5 - 1 1.5 - 1.45 t = 2 .303 × 10 3 = y × 10 3 ∴ y = 2.303 After round off: y = 2 .3

Question 12

In the decay sequence: U 92 238 &#8594; - x 1 T h 90 234 &#8594; - x 2 P a 91 234 &#8594; - x 3 Z &#160; 234 &#8594; - x 4 T h 90 234 x 1 , &#160; x 2 , &#160; x 3 and x 4 are the particles/ radi

Accepted Answer

I.e., x 1 = α , x 2 = β , x 3 = β and x 4 = α .

Question 13

Consider the kinetic data given in the following table for the reaction A + B + C &#8594; Product.   Experiment No. A m o l &#160; d m - 3 B m o l &#160; d m - 3 C m o l &#160; d m - 3 Rate of re

Accepted Answer

r = k [ A ] a [ B ] b [ C ] c 6 × 10 - 5 = k ( 0.2 ) a ( 0.1 ) b ( 0.1 ) c ..... (1) 6 × 10 - 5 = k ( 0.2 ) a ( 0.2 ) b ( 0.1 ) c ...... (2) 1.2 × 10 - 4 = k ( 0.2 ) a ( 0.1 ) b ( 0.2 ) c ...... (3) 9 × 10 - 5 = k ( 0.3 ) a ( 0.1 ) b ( 0.1 ) c ...... (4) From equation (1) and (ii) : b = 0 From equation (1) and (iii) : c = 1 From equation (1) and (iv) : a = 1 Put x , y and z in equation (i) we get; 6 × 10 - 5 = k ( 0.2 ) 1 ( 0.1 ) 0 ( 0.1 ) 1 r = 3 × 10 - 3 [ A ] [ C ] For [ A ] = 0.15 mol dm - 3

Chemical Kinetics — Previous Year Questions & Solutions