The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :

Question

The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :

The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :
A. `+46.0 kJ`
B. `+92.0 kJ`
C. `-23 kJ`
D. `-92 kJ`

Monjur Alahi

8 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
The given reaction involve decomposition of `2` mol of ammonia `DeltaH_(f)(NH_(3))=-46.0`, `DeltaH_("decomposition")"of" NH_(3)=+46`
`:. DeltaH "of given reaction"=2xx46=92 kJ`

8 views Answered 2023-02-05 15:29