हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Advertisements
Advertisements

प्रश्न

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

The rate of reaction becomes four times when the temperature changes from 293 K to 313 K.

Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature.

(R = 8.314 J K–1 mol–1)

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature.

[Given: log 4 = 0.602, log 2 = 0.301, R = 8.314 J K–1mol]

संख्यात्मक
Advertisements

उत्तर

Given: T1 = 293 K

T2 = 313 K

k2 = 4 × k1

R = 8.314 × 10−3 kJ K−1 mol−1

According to the Arrhenius equation,

`log_10  k_2/k_1 = E_a/(2.303 R) [1/T_1 - 1/T_2]`

⇒ `log_10  (4 xx k_1)/k_1 = E_a/(2.303 xx 8.314 xx 10^-3) xx [1/293 - 1/313]`

⇒ log10 (4) = `E_a/0.0191 xx (313 - 293)/(293 xx 313)`

⇒ 0.6021 = `E_a/0.0191 xx 20/(293 xx 313)`

⇒ Ea = `(0.6021 xx 0.0191 xx 293 xx 313)/20`

= `1054.66/20`

= 52.73 kJ mol−1

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Chemical Kinetics - Exercises [पृष्ठ ८८]

APPEARS IN

एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
अध्याय 3 Chemical Kinetics
Exercises | Q 3.30 | पृष्ठ ८८
नूतन Chemistry [English] Class 12 ISC
अध्याय 3 Chemical Kinetics
'NCERT TEXT-BOOK' Exercises | Q 4.30 | पृष्ठ २८३
नूतन Chemistry [English] Class 12 ISC
अध्याय 3 Chemical Kinetics
QUESTIONS FROM ISC EXAMINATION PAPERS | Q 19. | पृष्ठ २८४

संबंधित प्रश्न

The rate constant for the first-order decomposition of H2O2 is given by the following equation:

`logk=14.2-(1.0xx10^4)/TK`

Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes.

(Given: R = 8.314 JK–1 mol–1)


The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?


Consider a certain reaction \[\ce{A -> Products}\] with k = 2.0 × 10−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−1.


The decomposition of hydrocarbon follows the equation k = `(4.5 xx 10^11 s^-1) e^(-28000 K//T)`

Calculate Ea.


The decomposition of A into product has value of k as 4.5 × 103 s−1 at 10°C and energy of activation 60 kJ mol−1. At what temperature would k be 1.5 × 104 s−1?


In the Arrhenius equation for a first order reaction, the values of ‘A’ of ‘Ea’ are 4 × 1013 sec−1 and 98.6 kJ mol1 respectively. At what temperature will its half life period be 10 minutes?

[R = 8.314 J K1 mol1]


Explain the following terms :

Half life period of a reaction (t1/2)

 

 

 Predict the main product of the following reactions:


The chemical reaction in which reactants require high amount of activation energy are generally ____________.


Consider figure and mark the correct option.


During decomposition of an activated complex:

(i) energy is always released

(ii) energy is always absorbed

(iii) energy does not change

(iv) reactants may be formed


Which of the following statements are in accordance with the Arrhenius equation?

(i) Rate of a reaction increases with increase in temperature.

(ii) Rate of a reaction increases with decrease in activation energy.

(iii) Rate constant decreases exponentially with increase in temperature.

(iv) Rate of reaction decreases with decrease in activation energy.


The reaction between \[\ce{H2(g)}\] and \[\ce{O2(g)}\] is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain.


Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.


Match the statements given in Column I and Column II

  Column I Column I
(i) Catalyst alters the rate of reaction (a) cannot be fraction or zero
(ii) Molecularity (b) proper orientation is not there always
(iii) Second half life of first order reaction (c) by lowering the activation energy
(iv) `e^((-E_a)/(RT)` (d) is same as the first
(v) Energetically favourable reactions (e) total probability is one are sometimes slow (e) total probability is one
(vi) Area under the Maxwell Boltzman curve is constant (f) refers to the fraction of molecules with energy equal to or greater than activation energy

The activation energy in a chemical reaction is defined as ______.


The slope of Arrhenius Plot `("In"  "k"  "v"//"s" 1/"T")` of first-order reaction is −5 × 103 K. The value of Ea of the reaction is. Choose the correct option for your answer. [Given R = 8.314 JK−1mol−1]


Arrhenius equation can be represented graphically as follows:

The (i) intercept and (ii) slope of the graph are:


A schematic plot of ln Keq versus inverse of temperature for a reaction is shown below

The reaction must be:


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×