Question

What fraction of molecules in a gas at 300k collide with a energy equal to activation energy of 50 kj/mol

Answer 1

Answer:

In the given case:

E  

a

​  

=209.5kJ mol  

−1

=209500J mol  

−1

 

T=581K

R=8.314J K  

−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

 

ln  

x

=−E  

a

​  

/RT

logx=−  

2.303RT

E  

a

​  

 

​  

 

logx=−  

2.303×8.314JK  

−1

mol  

−1

×581

209500Jmol  

−1

 

​  

 =18.8323

Now, x= Anti log(18.8323)

=Antilog  

19

ˉ

.1677

Answer 2

$\huge\bf\boxed{\boxed{\underline{\orange{Question!!}}}}$

what fraction of molecules in a gas at 300k collides with an energy equal to activation energy of 50KJ/mol

$\huge\bf\boxed{\boxed{\underline{\orange{Answes!!}}}}$

$\huge\bold{Given}$

$\mathrm\red{Activation \:Energy \:Ea = 50KJ/mol^{-1}}$

$\mathrm\red{→ 50 × 10³ J/mol^{-1}}$

$\mathrm\blue{Temperature \:(T) = 300K}$

$\huge\bold{To \:find}$

Fraction of molecules with Energy equal to Ea

We know the formula :

$\huge\mathrm\purple{f = e^\frac{-Ea}{RT}}$

$\huge\mathrm\red{f = e^\frac{-50×10³mol^{-1}}{8.314JK^{-1}mol^{-1}×300K}}$

$\mathrm\pink{log10f = \frac{-50×10³}{8.314×300×2.303}}$

$\mathrm\pink{log10f = -8.70}$

$\mathrm\pink{f = antilog(-8.70)}$

$\mathrm\pink{f = 1.99 × 10^{-9}}$

$\mathrm\pink{f = 2.0 × 10^{-9}}$

Fraction of molecules with Energy equal to Ea = $2.0 × 10^{-9}$

$\huge\bf\boxed{\boxed{\underline{\blue{Ea = 2.0 × 10^{-9}}}}}$