wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Calculate the standard free energy change for the reaction,
N2(g)+3H2(g)2NH3(g) at 298 K.
Given Ho=92.4 kJmol1 and So=198.3 JK1mol1

A
-85.50 kJ/mol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
33.31 kJ/mol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
-33.31 kJ/mol
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
85.50 kJ/mol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C -33.31 kJ/mol
Go=HoTSo
=92.4298×(198.3)×103(So=198.3×103 kJK1mol1)
=33.306 kJ/mol
Since Go is negative, it means that a mixture of H2 and N2 at 25oC, each present at a pressure of 1 atm would react spontaneously to form ammonia.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon