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  3. Chemistry
  4. If the enthalpy change for the reaction is 2NH3(g)arrow N2(g)+3H2(g) is 92 kJ mol-1 the enthalpy of formation of ammonia is:

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Q. If the enthalpy change for the reaction is $ 2NH_{3}(g)\rightarrow N_{2}(g)+3H_{2}(g) $ is $ 92\,kJ\,mol^{-1} $ the enthalpy of formation of ammonia is:

J & K CETJ & K CET 2001

Solution:

$2 NH _{3}(g) \rightarrow N _{2}(g)+3 H _{2}(g) ; \Delta H =+92 \,kJ$
$N _{2}(g)+3 H _{2}(g) \rightarrow 2 NH _{3}(g) ; \Delta H =-92 \,kJ$
$\frac{1}{2} N _{2}(g)+\frac{3}{2} H _{2}(g) \rightarrow NH _{3}(g) ; \Delta H =-46\, kJ$
Hence, enthalpy of formation of $NH _{3}$ is $-46 \,kJ \,mol ^{-1}$