Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6

Q.

The bond dissociation enthalpies of H2(g),Cl2(g) and HCl(g) are 435, 243 and 431 kJ mol1 respectively. The enthalpy of formation of HCl(g) will be

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

770 kJ mol1

b

1109 kJ mol1

c

247 kJ mol1

d

92 kJ mol1

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

H2(g)+Cl2(g)2HCl(g)ΔfH(HCl,g)=12[ε(HH)+ε(ClCl)2ε(HCl)]=12(435+2432×431)kJ mol1=92 kJ mol1

Watch 3-min video & get full concept clarity

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon