मराठी

What is the Effect of Adding a Catalyst on Activation Energy (Ea)

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प्रश्न

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

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उत्तर

The catalyst provides an alternate pathway or reaction mechanism by reducing the activation energy (Ea) between reactants and products.

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2016-2017 (March) All India Set 3

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संबंधित प्रश्‍न

 

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


Activation energy of a chemical reaction can be determined by ______.


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.


Oxygen is available in plenty in air yet fuels do not burn by themselves at room temperature. Explain.


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


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]


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

The reaction must be:


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]


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