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

During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is:

A
2C4H10(g)+13O2(g)8CO2(g)+10H2O(l)ΔcH=2658.0kJmol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
C4H10(g)+132O2(g)4CO2(g)+5H2O(g)ΔcH=1329.0kJmol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
C4H10(g)+132O2(g)4CO2(g)+5H2O(l)ΔcH=2658.0kJmol1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
C4H10(g)+132O2(g)4CO2(g)+5H2O(l)ΔcH=+2658.0kJmol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C C4H10(g)+132O2(g)4CO2(g)+5H2O(l)ΔcH=2658.0kJmol1
Solution:- (C) C4H10(g)+132O2(g)4CO2(g)+5H2O(l);ΔcH=2658.0kJ/mol
The complete combustion of 1 mole of butane is represented by
C4H10(g)+132O2(g)4CO2(g)+5H2O(l);ΔcH=2658.0kJ/mol
ΔcH should be negative and have a value of 2658kJ/mol.

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