Question

The enthalpy of formation of ammonia gas at 298 K is - 46.11 kJ ${\rm{mo}}{{\rm{l}}^{ - 1}}$. The equation to which this value relates is:

Accepted Answer

$\frac{1}{2}{{m{N}}_2}\left( {m{g}} ight) + \frac{3}{2}{{m{H}}_2}\left( {m{g}} ight) 	o {m{N}}{{m{H}}_3}\left( {m{g}} ight)$

Answer

${m{N}}\left( {m{g}} ight) + 3{m{H}}\left( {m{g}} ight) 	o {m{N}}{{m{H}}_3}\left( {m{g}} ight)$

Answer

${{m{N}}_2}\left( {m{g}} ight) + 3{{m{H}}_2}\left( {m{g}} ight) 	o 2{m{N}}{{m{H}}_3}\left( {m{g}} ight)$

Answer

$\frac{1}{2}{{m{N}}_2}\left( {m{g}} ight) + \frac{3}{2}{{m{H}}_2}\left( {m{g}} ight) 	o {m{N}}{{m{H}}_3}\left( l ight)$

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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