CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The enthalpies of formation of OH(g), H(g) and O(g) are 21 kJ mol−1, 109 kJ mol−1and 124 kJ mol−1. The value of bond enthalpy O−H is

A
424 kJ mol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
212 kJ mol1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
424 kJ mol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
212 kJ mol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 212 kJ mol1
The enthalpy of formations of the given gases are :
fH(H,g)=109 kJ mol1
fH(O,g)=124 kJ mol1
fH(OH,g)=21 kJ mol1
We have to calculate H for the equation,
OH(g)H(g)+O(g)
H=fH(H,g)+fH(O,g)fH(OH,g)
H=(109+12421)kJ mol1H=212 kJ mol1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Thermochemistry
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon