Difficult
The activation energy for the reaction $2 HI_{(g)} \rightarrow H_{2(g)} + I_{2(g)}$ 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?
(N/A) In the given case:
$E_a = 209.5 \ kJ \ mol^{-1} = 209500 \ J \ mol^{-1}$
$T = 581 \ K$
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
The fraction of molecules of reactants having energy equal to or greater than activation energy is given by the Arrhenius factor $x = e^{-E_a / RT}$.
Taking the logarithm on both sides:
$\log x = -\frac{E_a}{2.303 \ RT}$
Substituting the values:
$\log x = -\frac{209500}{2.303 \times 8.314 \times 581}$
$\log x = -\frac{209500}{11124.5} \approx -18.8323$
Now,$x = \text{antilog}(-18.8323) = 10^{-18.8323} = 10^{0.1677} \times 10^{-19} \approx 1.471 \times 10^{-19}$.
$E_a = 209.5 \ kJ \ mol^{-1} = 209500 \ J \ mol^{-1}$
$T = 581 \ K$
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
The fraction of molecules of reactants having energy equal to or greater than activation energy is given by the Arrhenius factor $x = e^{-E_a / RT}$.
Taking the logarithm on both sides:
$\log x = -\frac{E_a}{2.303 \ RT}$
Substituting the values:
$\log x = -\frac{209500}{2.303 \times 8.314 \times 581}$
$\log x = -\frac{209500}{11124.5} \approx -18.8323$
Now,$x = \text{antilog}(-18.8323) = 10^{-18.8323} = 10^{0.1677} \times 10^{-19} \approx 1.471 \times 10^{-19}$.
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Some details of the above reaction are listed below.
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| Step | Rate constant $(s^{-1})$ | Activation energy $(kJ \ mol^{-1})$ |
|---|---|---|
| $1$ | $k_1$ | $300$ |
| $2$ | $k_2$ | $200$ |
| $3$ | $k_3$ | $Ea_3$ |
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