KCET · Chemistry · Thermodynamics (C)
The enthalpy of formation of \(\mathrm{NH}_{3}\) is \(-46 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The enthalpy change for the reaction
\(2 \mathrm{NH}_{3}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) is
- A \(+184 \mathrm{~kJ}\)
- B \(+23 \mathrm{~kJ}\)
- C \(+92 \mathrm{~kJ}\)
- D \(+46 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(C) \(+92 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
\(2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\)
\(\Delta \mathrm{H}_{\mathrm{r}}=-\left(2 \times\right.\) enthalpy of formation of \(\left.\mathrm{NH}_{3}\right)\)
\(\quad=-(2 \times-46)=92 \mathrm{~kJ}\)
\(\Delta \mathrm{H}_{\mathrm{r}}=-\left(2 \times\right.\) enthalpy of formation of \(\left.\mathrm{NH}_{3}\right)\)
\(\quad=-(2 \times-46)=92 \mathrm{~kJ}\)
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