The activation energy for the reaction `:` ltbr. `2Hl(g) rarr H_(2)(g)+I_(2)(g)` is `209.5 kJ mol^(-1)` at `581K`. Calculate the fraction of molecules

Question

The activation energy for the reaction : ltbr. 2Hl(g)→H2(g)+I2(g) is 209.5kJmol-1 at 581K . Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ? Select the correct answer from above options

Answer 1

Correct Answer - 1.47×1019 Fraction of molecules having energy equal to or greater than activation energy is given by : n N =(x)=e-Ea/RT ∴ln(x)= -Ea RT orlog(x)= -Ea 2.303RT = 209.5×Jmol-1 2.303×8.314JK-1mol-1×581K =-18.8323 ∴x=Antilog(-18.8323) =Antilog(-18.8323+1-1) =Antilog ¯ 19 .1677=1.471×10-19