Advertisements
Advertisements
प्रश्न
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)
Advertisements
उत्तर
According to the Arrhenius equation,
k=Ae(−Ea/RT)
From this, we get
`"log"k_2/k_1=E_a/(2.303R)((T_2-T_1)/(T_1T_2))`
We are given that
Initial temperature, T1=300 K
Final temperature, T2=310 K
Rate constant at initial temperature, k1=2×10−2
Rate constant at final temperature, k2=4×10−2
Gas constant, R=8.314 J K−1 mol−1
Substituting the values, we get
`"log"((4xx10^(-2))/(2xx10^(-2)))=E_a/(2.303xx8.314)((310-300)/(300xx310))`
`therefore " activation energy of the reaction, "E_a=(log2xx2.303xx8.314xx300xx310)/10`
`=535985.94" J mol"^(-1)`
`=535.98" kJ mol"^(-1)`
संबंधित प्रश्न
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 of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
Define activation energy.
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 ?
Predict the main product of the following reactions:
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.
Oxygen is available in plenty in air yet fuels do not burn by themselves at room temperature. Explain.
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 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.
A schematic plot of ln Keq versus inverse of temperature for a reaction is shown below
The reaction must be:
