English
Karnataka Board PUCPUC Science 2nd PUC Class 12

Why in the redox titration of KMnOX4 vs oxalic acid, we heat oxalic acid solution before starting the titration? - Chemistry

Advertisements
Advertisements

Question

Why in the redox titration of \[\ce{KMnO4}\] vs oxalic acid, we heat oxalic acid solution before starting the titration?

Short/Brief Note
Advertisements

Solution

The reaction between \[\ce{KMnO4}\] and oxalic acid is very slow. By raising the temperature we can increase the rate of reaction.

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Chemical Kinetics - Exercises [Page 56]

APPEARS IN

NCERT Exemplar Chemistry [English] Class 12
Chapter 4 Chemical Kinetics
Exercises | Q III. 49. | Page 56

RELATED QUESTIONS

 

Consider the reaction

`3I_((aq))^-) +S_2O_8^(2-)->I_(3(aq))^-) + 2S_2O_4^(2-)`

At particular time t, `(d[SO_4^(2-)])/dt=2.2xx10^(-2)"M/s"`

What are the values of the following at the same time?

a. `-(d[I^-])/dt`

b. `-(d[S_2O_8^(2-)])/dt`

c. `-(d[I_3^-])/dt`

 

 

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 of a first order reaction increases from 2 × 10−2 to 4 × 10−2 when the temperature changes from 300 K to 310 K. Calculate the energy of activation (Ea).

(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)


The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.


Calculate activation energy for a reaction of which rate constant becomes four times when temperature changes from 30 °C to 50 °C. (Given R = 8.314 JK−1 mol−1). 


What is the effect of adding a catalyst on Activation energy (Ea)


The decomposition of a hydrocarbon has value of rate constant as 2.5×104s-1 At 27° what temperature would rate constant be 7.5×104 × 3 s-1if energy of activation is  19.147 × 103 J mol-1 ?


 Write a condition under which a bimolecular reaction is kinetically first order. Give an example of  such a reaction. (Given : log2 = 0.3010,log 3 = 0.4771, log5 = 0.6990).


 Predict the main product of the following reactions:


The rate of chemical reaction becomes double for every 10° rise in temperature because of ____________.


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.


Mark the incorrect statements:

(i) Catalyst provides an alternative pathway to reaction mechanism.

(ii) Catalyst raises the activation energy.

(iii) Catalyst lowers the activation energy.

(iv) Catalyst alters enthalpy change of the reaction.


Why does the rate of a reaction increase with rise in temperature?


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

For an endothermic reaction energy of activation is Ea and enthalpy of reaction ΔH (both of there in KJ moI–1) minimum value of Ea will be ______.


In respect of the eqn k = \[\ce{Ae^{{-E_a}/{RT}}}\] in chemical kinetics, which one of the following statement is correct?


The equation k = `(6.5 xx 10^12 "s"^(-1))"e"^(- 26000 " K"//"T")` is followed for the decomposition of compound A. The activation energy for the reaction is ______ kJ mol-1. (Nearest integer) (Given: R = 8.314 JK-1 mol-1)


A first-order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction. [R = 8.314 J K−1 mol−1]

[Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]


It is generally observed that the rate of a chemical reaction becomes double with every 10°C rise in temperature. If the generalisation holds true for a reaction in the temperature range of 298 K to 308 K, what would be the value of activation energy (Ea) for the reaction?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×