Ncert Exercises & Solutions

Q. 13. Given

$N_{2 (g)} + 3 H_{2 (g)} \to 2 N H_{3 (g)}; ∆_{r} H^{θ} = - 92.4 k J m o l^{- 1}$

What is the standard enthalpy of formation of $N H_{3}$ gas?

NEETprep Answer:

Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.
Re-writing the given equation for 1 mole of

N H_{3 (g)}

\frac{1}{2} N_{2 (g)} + \frac{3}{2} H_{2 (g)} \to N H_{3 (g)}

∴

Standard enthalpy of formation of

N H_{3 (g)}

= \frac{1}{2} ∆_{r} H^{θ}

= \frac{1}{2} (- 92.4 k J m o l^{- 1})

= - 46.2 k J m o l^{- 1}

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