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

# The enthalpy of formation of ammonia is −46.0KJmol−1. The enthalpy change for the reaction:2NH3(g)→N2(g)+3H2(g)

A
46.0KJmol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
92.0KJmol1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
23.0KJmol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
92.0KJmol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 92.0KJmol−1We know that the enthalpy of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.Now the enthalpy of formation for ammonia is given to be.−46.0KJmol−1.for the following equation:12N2(g)+32H2(g)→NH3(g)→(1) Now in the give equation, the number of moles of ammonia is 2 and the reaction in equation (1) is reversed. Hence, enthalpy of formation of the given reaction is =2×46.0 KJmol−1=92KJmol−1 Hence, option (b) is the correct answer.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Hydrides of Nitrogen - Ammonia and Hydrazine
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program