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 Multiple Choice QuestionsMultiple Choice Questions

1.

The resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution of 0.5 M solution of the same electrolyte is 1.4 S m-1 and resistance of the same solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in Sm-2 mol-1 is

  • 5 x 10-4

  • 5 x 10-3

  • 5 x 103

  • 5 x 102


A.

5 x 10-4

For first solution,
k = 1.4 Sm-1, R =50 Ω , M =0.2
specific conductance
straight k space equals space 1 over straight R space straight x l over straight A
1.4 space Sm to the power of negative 1 end exponent space equals space 1 over 50 straight x l over straight A
l over straight A space equals space 50 space straight x space 1.4 space straight m to the power of negative 1 end exponent
straight k space equals 1 over 280 space straight x space 1.4 space straight x space 50 space equals space 1 fourth
Now comma space molar space conductivity
straight lambda subscript straight m space equals space fraction numerator straight k over denominator 1000 space xm end fraction space equals space fraction numerator 1 divided by 4 over denominator 1000 straight x 0.5 end fraction space equals space 1 over 2000
space equals space 5 space straight x space 10 to the power of negative 4 end exponent space Sm squared mol to the power of negative 1 end exponent

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2.

Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to-
(R = 8.314 J mol–1K–1)

  • 8

  • 12

  • 6

  • 4


D.

4

From Arrhenius equation

straight K space equals space straight A. straight e to the power of fraction numerator negative Ea over denominator RT end fraction end exponent
So comma space straight K subscript 1 space equals space straight A. straight e to the power of bevelled fraction numerator negative straight E subscript straight a 1 end subscript over denominator RT end fraction end exponent space..... space left parenthesis 1 right parenthesis
straight K subscript 2 space equals space straight A. straight e to the power of bevelled fraction numerator negative straight E subscript straight a 2 end subscript over denominator RT end fraction end exponent space..... space left parenthesis 2 right parenthesis
so space divide space equation space 2 space by space 1 space we space get
rightwards double arrow space straight K subscript 1 over straight K subscript 2 space equals space straight e to the power of fraction numerator left parenthesis straight E subscript straight a 1 end subscript minus straight E subscript straight a 2 end subscript over denominator RT end fraction end exponent
Taking space log space on space both space side
ln space open parentheses straight K subscript 2 over straight K subscript 1 close parentheses space equals space fraction numerator straight E subscript straight a 1 end subscript minus straight E subscript straight a 2 end subscript over denominator RT end fraction space equals space fraction numerator 10 comma 000 over denominator 8.314 space straight x space 300 end fraction space equals space 4

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3.

The rate of a chemical reaction doubles for every 10ºC rise of temperature. If the temperature is raised by 50ºC, the rate of the reaction increases by about

  • 10 times

  • 24 times

  • 32 times

  • 64 times


C.

32 times

For every 10o C rise of temperature, the rate is doubled. Thus, temperature coefficient of the reaction = 2 when temperature is increased by 50o rate becomes
fraction numerator Rate space at space 50 degree space straight C over denominator Rate space at space straight T subscript 1 superscript degree space straight C end fraction space equals space left parenthesis 2 right parenthesis to the power of fraction numerator increment straight T over denominator straight T subscript 1 end fraction end exponent space equals space space left parenthesis 2 right parenthesis to the power of 50 over 10 end exponent space equals space 2 to the power of 5 space
space equals space 32 space times

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4.

Higher order (>3) reactions are rare due to:

  • the increase in entropy and activation energy as more molecules are involved.

  • shifting of equilibrium towards reactants due to elastic collisions

  • loss of active species on a collision

  • low probability of simultaneous collision of all the reacting species


A.

the increase in entropy and activation energy as more molecules are involved.

Conditions for the occurrence of a reaction:

(i)  Proper orientation and effective collision of the reactants.
(ii)  the chances of simultaneous collision with proper orientation between more than 3 species are very rare, so reaction with order greater than 3 are rare.

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5.

Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:

  • 6.93×10−4 mol min−1

  • 2.66 L min−1 at STP

  • 1.34×10−2 mol min−1

  • 6.93×10−2 mol min−1


A.

6.93×10−4 mol min−1

straight k space equals space open parentheses fraction numerator 0.693 minus 1 over denominator 25 end fraction close parentheses space min to the power of negative 1 end exponent

or space fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction space equals space minus 1 half fraction numerator straight d left square bracket straight H subscript 2 straight O subscript 2 right square bracket over denominator dt end fraction
space equals space fraction numerator straight k left square bracket straight H subscript 2 straight O subscript 2 right square bracket over denominator 2 end fraction space equals 6.93 space straight x space 10 to the power of negative 4 end exponent space mol space min to the power of negative 1 end exponentFor first order reaction

straight k space equals fraction numerator 2.303 over denominator straight t end fraction log space fraction numerator straight a over denominator straight a minus straight x end fraction
Given colon
straight t equals space 50 space min comma
straight a equals space 0.5 straight M
straight a minus straight x space equals space 0.125 space straight M

therefore space straight k space equals space fraction numerator 2.303 over denominator 50 end fraction space log fraction numerator 0.5 over denominator 0.125 end fraction space equals space 0.0277 space min to the power of negative 1 end exponent
Now comma space as space per space reaction

2 straight H subscript 2 straight O subscript 2 space rightwards arrow with space on top space 2 straight H subscript 2 straight O space plus space straight O subscript 2
fraction numerator negative 1 over denominator 2 end fraction fraction numerator straight d left square bracket straight H subscript 2 straight O subscript 2 right square bracket over denominator dt end fraction space equals space 1 half fraction numerator straight d left square bracket straight H subscript 2 straight O right square bracket over denominator dt end fraction space equals space fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction
Rate space of space reaction comma space
fraction numerator negative straight d left square bracket straight H subscript 2 straight O subscript 2 right square bracket over denominator dt end fraction space equals space straight k left square bracket straight H subscript 2 straight O subscript 2 right square bracket
therefore space fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction space equals space minus space 1 half fraction numerator straight d left square bracket straight H subscript 2 straight O subscript 2 right square bracket over denominator dt end fraction space equals space 1 half straight k left square bracket straight H subscript 2 straight O subscript 2 right square bracket space... space left parenthesis straight i right parenthesis
When space the space concentration space of space straight H subscript 2 straight O subscript 2 space reaches space 0.05 space straight M comma
fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction space equals space 1 half space straight x space 0.0277 space straight x space 0.05 space space space left square bracket from space equ space left parenthesis straight i right parenthesis right square bracket
or space
fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction space equals space 6.93 space straight x space 10 to the power of negative 4 end exponent space mol space min to the power of negative 1 end exponent

Alternative Method:
If fifty minutes, the concentration of H2O2 decreases from 0.5 to 0.125 M or in one half-life, concentration of H2O2 decreases from 0.5 to 0.25 M. In two half-lives, concentration of H2O2 decreases from 0.5 to 0.125 M or 2t1/2 = 50 min
t1/2 = 25 min
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6.

The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be (R = 8.314 JK–1 mol–1 and log 2 = 0.301)

  • 53.6 kJ mol-1

  • 48.6 kJ mol-1

  • 58.5 kJ mol-1

  • 60.5 kJ mol-1


A.

53.6 kJ mol-1

By using Arrhenius equation,

log space straight k subscript 2 over straight k subscript 1 space equals space fraction numerator negative straight E subscript straight a over denominator 2.303 space straight R end fraction open parentheses 1 over straight T subscript 2 minus 1 over straight T subscript 1 close parentheses space.... space left parenthesis straight i right parenthesis
therefore space straight k subscript 1 over straight k subscript 2 space equals space straight r subscript 1 over straight r subscript 2
rightwards double arrow straight k subscript 1 over straight k subscript 2 space equals space fraction numerator straight r subscript 1 over denominator 2 straight r subscript 1 end fraction
straight k subscript 1 over straight k subscript 2 space equals space 1 half
straight k subscript 2 over straight k subscript 1 space equals space 2
Given, T2 = 310; T1 = 300K
On putting values in Eq (i), we get
log space 2 space equals space fraction numerator negative straight E subscript straight a over denominator 2.303 space straight x space 8.314 end fraction open parentheses 1 over 310 minus 1 over 300 close parentheses
rightwards double arrow space straight E subscript straight a space equals space 53598.6 space straight J divided by mol space equals space 53.6 space kJ divided by mol

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7.

For a first order reaction, (A) → products, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is

  • 1.73 x 10–5 M/ min

  • 3.47 x 10–4 M/min

  • 3.47 x 10–5 M/min

  • 1.73 x 10–4 M/min


B.

3.47 x 10–4 M/min

By first order kinetic rate constant,

straight k space equals space fraction numerator 2.303 over denominator straight t end fraction space log open parentheses fraction numerator straight a over denominator straight a minus straight x end fraction close parentheses
straight a space equals space 0.1 space straight M
left parenthesis straight a minus straight x right parenthesis space equals space 0.025 space straight M comma space straight t space equals space 40 space min
straight k space equals space fraction numerator 2.303 over denominator 40 end fraction space log space fraction numerator 0.1 over denominator 0.025 space straight M end fraction
equals space 0.0347 space min to the power of negative 1 end exponent
Rate space equals space open parentheses dx over dt close parentheses space equals space straight k left square bracket straight A right square bracket to the power of 1
space equals space 0.0347 space straight x space 0.01
equals space 3.47 space straight x space 10 to the power of negative 4 end exponent space straight M space min to the power of negative 1 end exponent

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8.

Rate law for the reaction, A+ B → product is rate = k[A]2[B] What is the rate constant; if rate of reaction at a given temperature is 0.22 Ms-1, when [A]= 1 M band [BJ= 0.25 M?

  • 3.52 M-2s-1

  • 0.88  M-2s-1

  • 1.136  M-2s-1

  • 0.05  M-2s-1


B.

0.88  M-2s-1

 For reaction; A+ B product dxdt= k [A]2[B]     0.22 = k(1)2 (0.25) k =0.220.25=0.88 M-2s-1


9.

The time for half life period of a certain reaction A → Products is 1 hour. When the initial concentration of the reactant ‘A’ is 2.0 mol L–1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L–1 if it is a zero order reaction?

  • 1 h 

  • 4 h 

  • 0.25 h 

  • 0.5 h


C.

0.25 h 

Given that [A]o = 2 mol L-1
straight t subscript 1 divided by 2 end subscript space equals space fraction numerator left square bracket straight A subscript 0 right square bracket over denominator 2 straight k subscript 0 end fraction space or comma space straight k subscript 0 space equals fraction numerator space left square bracket straight A subscript 0 right square bracket over denominator 2 straight t subscript 1 divided by 2 end subscript end fraction
On space putting space the space value comma space we space get
straight k subscript 0 space equals space fraction numerator left square bracket 2 right square bracket over denominator 2 space straight x 1 end fraction space equals space 1
So comma space straight k subscript 0 space equals 1
and space straight k subscript 0 space equals space fraction numerator increment straight X over denominator straight t end fraction
or space straight t space equals space fraction numerator 0.50 minus 0.25 over denominator 1 end fraction space equals space 0.25 space straight h

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10.

For the non- stoichiometric reaction 2A + B → C + D, the following kinetic data were obtained in three separate experiment, all at 298 K.

  Initial concentration (A) Initial concnetration (B) Initial rate of formation of C (mol L-1 S-1)
1 0.1 M 0.1 M 1.2 x 10-3
2 0.1 M 0.2 M 1.2 x 10-3
3 0.2 M 0.1 M 2.4 x 10-3
 The rate law for the formation of C is
  • dC over dt space equals space straight k space left square bracket straight A right square bracket left square bracket straight B right square bracket
  • dC over dt space equals space straight k open square brackets straight A squared close square brackets left square bracket straight B right square bracket
  • dC over dt space equals space straight k left square bracket straight A right square bracket left square bracket straight B right square bracket squared
  • dC over dt equals space straight k space left square bracket straight A right square bracket

D.

dC over dt equals space straight k space left square bracket straight A right square bracket
straight r space equals space dC over dt space equals space straight k space left square bracket straight A right square bracket to the power of straight x left square bracket straight B right square bracket to the power of straight y
where x = order of reaction w.r.t A
 y = order of reaction w.r.t B
1.2 x 10-3 = k (0.1)x (0.1)y
1.2 x 10-3 = k (0.1)x (0.2)y
2.4 x 10-3 = k(0.2)x (0.1)y
R = k [A]1[B]0

As shown above, rate of reaction remains constant as the concentration of  reactant (B) changes from 0.1 M to 0.2 M and becomes double when concentration of A change from 0.1 to 0.2
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