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The heat change for the reaction, N(2)...

The heat change for the reaction,
`N_(2)(g)+3H_(2)(g)to2NH_(3)(g)`
is `-92.2kJ`. Calculate the heat of formation of ammonia.

Text Solution

Verified by Experts

The enthalpy of formation of ammonia is the enthalpy change for the formation of `1` mole of ammonia from its elements, i.e.,
`1/2 N_(2)(g) + 3/2 H_(2)(g) to NH_(3)(g), Delta_(f)H^(@) = ?`
The enthalpy change for the reaction
`N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g)`
is `Delta_(r)H = -92.2 kJ`.
This equation corresponds to formation of two moles of ammonia. Thus,
`Delta_(f) H^(@) = -(92.2)/(2) = -46.1 kJ mol^(-1)`
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