Question

The activation energy for the reaction 2HI(g)​→H2​I2(g)​ is 209.5mol^−1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy. 

Answer 1

Answer:

$1.471*10^-^1^9$

Explanation:

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

$x=e^-^E^a^/^R^T\\In\ x= -Eₐ/RT$

$logx=-\frac{Eₐ}{2.303\ RT}$

$\boxed{logx=\frac{209500\ J\ mol^-^1}{2.303*8.314JK^-^1\ mol^-^1*581} =18.8323}$

$Now,\ x=Anti\ log(18.8323)\\\\=Anti\ log\ 19(bar).1677\\\\=1.471*10^-^1^9$

Answer 2

Answer:

hope this helps.