Ncert Exercises & Solutions

Question 4.9: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:

$E_{a} = 209.5 k j m o l - 1 = 209500 k m o l - 1$
$T = 581 K$
$R = 8.314 j k^{- 1}$
Now, the fraction of molecules of reactants having energy equal to or greater than

activation energy is given as:

$x = e^{- E a / R T}$
$\Rightarrow \ln x = - E_{a} / R T$
$\Rightarrow \log x = - \frac{E_{a}}{203.0 R T}$
$\Rightarrow \log x = \frac{209500 j m o l^{- 1}}{2.303 \times 8.314 j k^{- 1} m o l^{- 1} \times 581}$
$N o w, x = A n t i \log (18.82323)$
$= A n t i \log \bar{19 .} 1677$
$= 1.471 \times 10^{- 19}$

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