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  4. The enthalpy of formation of ammonia is -46.0 kJ mol -1 . The enthalpy for the reaction 2 N 2(g)+6 H 2(g) arrow 4 NH 3( g ) is equal to -

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Q. The enthalpy of formation of ammonia is $-46.0 \,kJ$ $mol ^{-1} .$ The enthalpy for the reaction $2 N _{2}(g)+6 H _{2}(g) \rightarrow$ $4 NH _{3}( g )$ is equal to -

Thermodynamics

Solution:

$2 N _{2}( g )+6 H _{2}( g ) \rightarrow 4 NH _{3}( g )$

$\Delta H =-46 \,kJ / mole$

Formation of $1\,. mole\, NH _{4}=46\, kJ / mole$

4 mole $NH _{4}=-4 \times 4.6$

$=-184 \,kJ / mole$