मराठी

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

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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

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


The rate constant for the decomposition of N2O5 at various temperatures is given below:

T/°C 0 20 40 60 80
105 × k/s−1 0.0787 1.70 25.7 178 2140

Draw a graph between ln k and `1/T` and calculate the values of A and Ea. Predict the rate constant at 30º and 50ºC.


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?


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). 


 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).


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.


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


Explain how and why will the rate of reaction for a given reaction be affected when the temperature at which the reaction was taking place is decreased.


The activation energy of one of the reactions in a biochemical process is 532611 J mol–1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300 = x × 10–3 k310. The value of x is ______.

[Given: ln 10 = 2.3, R = 8.3 J K–1 mol–1]


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