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What will be the effect of temperature on rate constant?
The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,
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Where,
A is the Arrhenius factor or the frequency factor
T is the temperature
R is the gas constant
Ea is the activation energy
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
It is given that T1 = 298 K
∴T2 = (298 + 10) K
= 308 K
We also know that the rate of the reaction doubles when temperature is increased by 10°.
Therefore, let us take the value of k1 = k and that of k2 = 2k
Also, R = 8.314 J K−1 mol−1
Now, substituting these values in the equation:
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We get:
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_13-11-08_Utpal_12_Chemistry_4_9_html_782090af.gif)
= 52897.78 J mol−1
= 52.9 kJ mol−1
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
The activation energy for the reaction 2HI(g)→ H2 + 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?
In the given case:
Ea = 209.5 kJ mol−1 = 209500 J 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=e−Ea/RT⇒In x=−EaRT⇒log x=−Ea2.303RT⇒log x=−209500 J mol−12.303×8.314 JK−1mol−1×581=−18.8323Now, x =Antilog (−18.8323) = 1.471×10−19
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