MHT CET · Chemistry · Thermodynamics (C)
From the given reaction, \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})} \Delta \mathrm{H}=-92 \cdot 6 \mathrm{~kJ}\), the enthalpy of formation of \(\mathrm{NH}_{3}\) is
- A \(-92 \cdot 6 \mathrm{~kJ}\)
- B \(-138 \cdot 9 \mathrm{k} \mathrm{J}\)
- C \(-185 \cdot 2 \mathrm{~kJ}\)
- D \(-46 \cdot 3 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(D) \(-46 \cdot 3 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
(B)
\(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})} \quad \Delta \mathrm{H}=-92.6 \mathrm{~kJ}\) \(\therefore\) Enthalpy of formation of \(\mathrm{NH}_{3}=\frac{-92.6}{2}=-46.3 \mathrm{~kJ}\)
\(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})} \quad \Delta \mathrm{H}=-92.6 \mathrm{~kJ}\) \(\therefore\) Enthalpy of formation of \(\mathrm{NH}_{3}=\frac{-92.6}{2}=-46.3 \mathrm{~kJ}\)
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