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प्रश्न
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.
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उत्तर
The rate constants for the decomposition of N2O5 at different temperatures are shown below.
| T(°C) | T(K) | `1/T` | k(s−1) | ln k (= 2.303 log k) |
| 0 | 273 | 3.6 × 10−3 | 7.87 × 10−7 | −14.06 |
| 20 | 293 | 3.4 × 10−3 | 1.70 × 10−5 | −10.98 |
| 40 | 313 | 3.19 × 10−3 | 25.7 × 10−5 | −8.266 |
| 60 | 333 | 3.00 × 10−3 | 178 × 10−5 | −6.332 |
| 80 | 353 | 2.8 × 10−3 | 2140 × 10−5 | −3.844 |
Slope of the line = tan θ
= `(y_2 - y_1)/(x_2 - x_1)`
= `(-10.98 - (-14.06))/(3.4 - 3.6) xx 10^3`
= `3.08/-0.2 xx 10^3`
= −15.4 × 103
Ea = −slope × R
= −(−15.4 × 103 × 8.314)
= 128.035 kJ K−1 mol−1
Again,
ln A = `ln k + E_a/(RT)`
= `-14.06 + (128.035 xx 10^3)/(8.314 xx 273)`
= `-14.06 + 128035/2269.722`
= −14.06 + 56.40
ln A = 42.34
⇒ A = antilog 42.34
= 0.3388 × 1019
Values of the rate constant k at 303 K and 323 K can be obtained from the graph.
First, k is obtained corresponding to `1/(303 K) and 1/(323 K)`, and then k is calculated.
संबंधित प्रश्न
Explain a graphical method to determine activation energy of a reaction.
(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`
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 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 activation energy for the reaction \[\ce{2 HI_{(g)} -> H2_{(g)} + I2_{(g)}}\] is 209.5 kJ mol−1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
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?
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)
Predict the main product of the following reactions:
Consider figure and mark the correct option.
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
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 |
The activation energy in a chemical reaction is defined as ______.
Arrhenius equation can be represented graphically as follows:
The (i) intercept and (ii) slope of the graph are:
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 ______.
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?
