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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
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Question : 39 of 39
Marks: +1, -0
The activation energy for the reaction 2HI(g)→H2(g)+I2(g)2\mathrm{HI}_{(g)} \rightarrow \mathrm{H}_{2(g)} + \mathrm{I}_{2(g)} is 209.5 kJ mol−1^{-1} at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
Solution:  
Fraction of molecules having energy equal to or greater thanactivation energy is given as
x=nN=e−EaRTx = \frac{n}{N} = e^{-\frac{E_{a}}{R T}}
∴ln⁡x=−EaRT\therefore \ln x = -\frac{E_{a}}{R T}
or, log⁡x=−Ea2.303RT\log x = -\frac{E_{a}}{2.303 R T}
or, log⁡x=−209.5×1032.303×8.314×581\log x = -\frac{209.5 \times 10^{3}}{2.303 \times 8.314 \times 581}
=−18.8323= -18.8323
∴    x=\therefore \;\; x = Antilog (−18.8323)(-18.8323)
=1.471×10−19= 1.471 \times 10^{-19}
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