Advertisements
Advertisements
प्रश्न
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
Advertisements
उत्तर
According to the Arrhenius equation,
`log "k"_2/"k"_1 = ("E"_"a")/(2.303 "R") [1/"T"_1 - 1/"T"_2]`
`"K"_2/"k"_1` = 2,
T1 = 298 K,
T2 = 308 K,
R = 8.314 JK−1 mol−1
∴ `log 2 = "E"_"a"/(2.303 xx 8.314 "JK"^-1 "mol"^-1) [1/(298 "K") - 1/(308 "K")]`
`0.3010 = "E"_"a"/(2.303 xx 8.314 "JK"^-1 "mol")^-1 xx 10/(298 xx 308)`
∴ `"E"_"a" = (0.3010 xx 2.303 xx 8.314 xx 298 xx 308)/10 "J mol"^-1`
= 52897.7 J mol−1
= 52.897 kJ mol−1
APPEARS IN
संबंधित प्रश्न
Explain a graphical method to determine activation energy of a reaction.
The rate constant of a first order reaction increases from 4 × 10−2 to 8 × 10−2 when the temperature changes from 27°C to 37°C. Calculate the energy of activation (Ea). (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
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)
The activation energy for the reaction \[\ce{2 HI_{(g)} -> H2_{(g)} + I2_{(g)}}\] is 209.5 kJ mol−1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
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 rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
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 ?
The rate of chemical reaction becomes double for every 10° rise in temperature because of ____________.
Consider figure and mark the correct option.
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
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.
Why does the rate of a reaction increase with rise in temperature?
Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
The activation energy in a chemical reaction is defined as ______.
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]
What happens to the rate constant k and activation energy Ea as the temperature of a chemical reaction is increased? Justify.
