Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
In the given case:
Ea = 209.5 kJ mol−1
T = 581 K
R = 8.314 JK−1 mol−1
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:
x = `"n"/"N"`
x = `e^((-E_a)/"RT")`
In x = `- (E_a)/"RT"`
or, log x = `-E_a/(2.303 "RT")`
or, log x = `-(209.5 xx 10^3 "J mol"^-1)/(2.303 xx 8.314 "JK"^-1 "mol"^-1 xx 581 "K")`
= −18.8323
x = antilog (−18.8323)
= antilog 19.1677
= 1.471 × 10−19
APPEARS IN
संबंधित प्रश्न
Explain a graphical method to determine activation energy of a reaction.
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 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?
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
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 ?
Consider figure and mark the correct option.
Match the statements given in Column I and Column II
| Column I | Column I | |
| (i) | Catalyst alters the rate of reaction | (a) cannot be fraction or zero |
| (ii) | Molecularity | (b) proper orientation is not there always |
| (iii) | Second half life of first order reaction | (c) by lowering the activation energy |
| (iv) | `e^((-E_a)/(RT)` | (d) is same as the first |
| (v) | Energetically favourable reactions (e) total probability is one are sometimes slow | (e) total probability is one |
| (vi) | Area under the Maxwell Boltzman curve is constant | (f) refers to the fraction of molecules with energy equal to or greater than activation energy |
In respect of the eqn k = \[\ce{Ae^{{-E_a}/{RT}}}\] in chemical kinetics, which one of the following statement is correct?
The rate constant for a reaction is 1.5 × 10–7 sec–1 at 50°C. What is the value of activation energy?
Arrhenius equation can be represented graphically as follows:
The (i) intercept and (ii) slope of the graph are:
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 decomposition of N2O into N2 and O2 in the presence of gaseous argon follows second-order kinetics, with k = (5.0 × 1011 L mol−1 s−1) `"e"^(-(29000 "K")/"T")`. Arrhenius parameters are ______ kJ mol−1.
An exothermic reaction X → Y has an activation energy 30 kJ mol-1. If energy change ΔE during the reaction is - 20 kJ, then the activation energy for the reverse reaction in kJ is ______.
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]
It is generally observed that the rate of a chemical reaction becomes double with every 10°C rise in temperature. If the generalisation holds true for a reaction in the temperature range of 298 K to 308 K, what would be the value of activation energy (Ea) for the reaction?
Which plot of ln k vs `1/T` is consistent with the Arrhenius equation?
