Chemical Kinetics

Question
CBSEENCH12010910

The photon of hard gamma radiation knocks a proton out of 2412Mg nucleus to form

  • the isotope of parent nucleus

  • the isobar of parent nucleus

  • the nuclide Na presubscript 11 presuperscript 23

  • the isobar of Na presubscript 11 presuperscript 23

Solution

C.

the nuclide Na presubscript 11 presuperscript 23

Question
CBSEENCH12010940

The rate equation for the reaction 2A + B → C is found to be: rate k[A][B]. The correct statement in relation to this reaction is that the

  • unit of K must be s-1

  • values of k is independent of the initial concentration of A and B

  • rate of formation of C is twice the rate of disappearance of A

  • t1/2 is a constant

Solution

B.

values of k is independent of the initial concentration of A and B

values of k is independent of the initial concentration of A and B

Sponsor Area

Question
CBSEENCH12010721

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

Solution

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

Sponsor Area

Question
CBSEENCH12010679

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

Solution

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

Question
CBSEENCH12010651

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

Solution

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