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प्रश्न
Oxygen is available in plenty in air yet fuels do not burn by themselves at room temperature. Explain.
उत्तर
The activation energy for combustion reactions of fuels is very high at room temperature therefore they do not burn by themselves.
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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 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 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?
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).
Explain the following terms :
Half life period of a reaction (t1/2)
Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
Arrhenius equation can be represented graphically as follows:
The (i) intercept and (ii) slope of the graph are:
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]
A schematic plot of ln Keq versus inverse of temperature for a reaction is shown below
The reaction must be:
