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प्रश्न
Why in the redox titration of \[\ce{KMnO4}\] vs oxalic acid, we heat oxalic acid solution before starting the titration?
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उत्तर
The reaction between \[\ce{KMnO4}\] and oxalic acid is very slow. By raising the temperature we can increase the rate of reaction.
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संबंधित प्रश्न
The rate constant of a first order reaction increases from 4 × 10−2 to 8 × 10−2 when the temperature changes from 27°C to 37°C. Calculate the energy of activation (Ea). (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
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?
The rate constant of a first order reaction are 0.58 S-1 at 313 K and 0.045 S-1 at 293 K. What is the energy of activation for the reaction?
Define activation energy.
Explain the following terms :
Half life period of a reaction (t1/2)
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
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.
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Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
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 ______.
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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:
An exothermic reaction X → Y has an activation energy 30 kJ mol-1. If energy change ΔE during the reaction is - 20 kJ, then the activation energy for the reverse reaction in kJ is ______.
A schematic plot of ln Keq versus inverse of temperature for a reaction is shown below
The reaction must be:
The rate of a reaction quadruples when temperature changes from 27°C to 57°C calculate the energy of activation.
(Given: R = 8. 314 J K−1 mol−1, log 4 = 0.6021)
