InterviewSolution
Saved Bookmarks
InterviewSolution
| 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 kJ`B. `64 kJ`C. `80 kJ`D. `92 kJ` |
|
Answer» Correct Answer - D The enthalpy of formation of ammonia describes the following thermochemical equation: `(1)/(2) N_(2) (g) + (3)/(2) H_(2) (g) rarr 2 NH_(3)` To get the enthalpy change for the given reaction, we need to reverse it and then multiply it by 2. Thus, `Delta H = - (2) (46.2) = 92.4 kJ` |
|