Question

He activation energy for the reaction, 2 hi(g) -> h2+i2 (g) is 209.5 k j mol-1 at 581 k.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Answer 1

Given that  

Activation energy, Ea = 209.5 kJ mol−1  

Multiply by 1000 to convert in j  

Ea= 209500 J mol−1  

Temperature, T = 581 K

Gas constant, R = 8.314 JK−1 mol−1

According to Arhenious equation

K = A e –Ea/RT  

In this formula term e –Ea/RT represent the number of molecules which have energy equal or more than activation energy  

Number of molecules = e –Ea/RT  

Plug the values we get  

Number of molecules

 

Find the value of 1 / anti ln(43.4 ) we get  

1.47 x 10-19

Do you know any other method of solving this problem ?

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?

Answer 2

Given,

Ka = 209.5KJ/mol = 209500J/mol

T=581K

Fraction of molecule of react and having energy equal to or greater than activation energy = e^{-Ka/RT}

or, e^{-209500/(8.314x581)}

or, e^{-43.4}

or, 1.41x10^(-19) molecules