

InterviewSolution
Saved Bookmarks
1. |
The enthalpy of formation of ammonia is `- 46.2 mol^(-1)`. The enthalpy change for the reaction `2NH_(3) rarr N_(2) + 3 H_(2)` isA. `42.0 kJ`B. `64.0 kJ`C. `80.0 kJ`D. `92.0 kJ` |
Answer» Correct Answer - D `underset(2"moles")(2NH_(3))toN_(2)+3H_(2)` The enthalpy of formation of ammonia `=-46kJ mol^(-1)` `DeltaH"of given reaction " =-2xx(-46)=92kJ` |
|