Q.

The enthalpy of formation of H2O(1) is - 290 kJ / mol and enthalpy of neutralization of strong acid with strong alkali is - 56 kJ / equiv. what is the enthalpy of formation of OH-  (aq)? Given Δform HH+,aq=0

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

- 334 kJ / mol

b

- 234 kJ / mol

c

- 346 kJ / mol

d

- 178 kJ / mol

answer is B.

(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)+12O2(g)→H2O(l); ΔH1=−290kJH++OH−→H2O; ΔH2=−56kJ12H2→H+(aq); ΔH=0∴ 12H2(g)+12O2(g)→OH−(aq)ΔrHΔrH=−290+56=−234kJ/mol
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon
The enthalpy of formation of H2O(1) is - 290 kJ / mol and enthalpy of neutralization of strong acid with strong alkali is - 56 kJ / equiv. what is the enthalpy of formation of OH-  (aq)? Given Δform HH+,aq=0