Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
The rate of reaction will decrease. At lower temperatures, the kinetic energy of molecules decreases thereby the collisions decrease resulting in a lowering of the rate of reaction.
APPEARS IN
संबंधित प्रश्न
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
`logk=logA-E_a/2.303R(1/T)`
Where Ea is the activation energy. When a graph is plotted for `logk Vs. 1/T` a straight line with a slope of −4250 K is obtained. Calculate ‘Ea’ for the reaction.(R = 8.314 JK−1 mol−1)
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)
What will be the effect of temperature on rate constant?
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).
The chemical reaction in which reactants require high amount of activation energy are generally ____________.
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.
Arrhenius equation can be represented graphically as follows:
The (i) intercept and (ii) slope of the graph are:
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
