Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

The activation energy for the reaction 2 HI(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?

The activation energy for the reaction 2 HI(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?

Grade:8

1 Answers

Gaurav
askIITians Faculty 164 Points
6 years ago

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−1mol−1
According to Arhenious equation
K = A e–Ea/RT
In this formula term e–Ea/RTrepresent the number of molecules which have energy equal or more than activation energy
Number of molecules = e–Ea/RT
Plug the values and solving we get
Number of molecules
1.47 x 10-19

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free