Advertisements
Advertisements
प्रश्न
Oxygen is available in plenty in air yet fuels do not burn by themselves at room temperature. Explain.
Advertisements
उत्तर
The activation energy for combustion reactions of fuels is very high at room temperature therefore they do not burn by themselves.
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)
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`
What will be the effect of temperature on rate constant?
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.
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.
In the Arrhenius equation for a first order reaction, the values of ‘A’ of ‘Ea’ are 4 × 1013 sec−1 and 98.6 kJ mol−1 respectively. At what temperature will its half life period be 10 minutes?
[R = 8.314 J K−1 mol−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).
A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given : log 2 = 0·3010, log 4 = 0·6021, R = 8·314 JK–1 mol–1)
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 ?
Activation energy of a chemical reaction can be determined by ______.
Which of the following graphs represents exothermic reaction?
(a)

(b)

(c)

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
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.
Why does the rate of a reaction increase with rise in temperature?
The rate constant for a reaction is 1.5 × 10–7 sec–1 at 50°C. What is the value of activation energy?
The activation energy in a chemical reaction is defined as ______.
The activation energy in a chemical reaction is defined as ______.
