Reaction for which enthalpy of reaction is equal to standard enthalpy of formation:

H_{2} (g) + C l_{2} (g) \to 2 H C l (g)

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H_{2} (g) + B r_{2} (l) \to 2 B r (l)

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C a O (s) + C O_{2} (g) \to C a C O_{3} (s)

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

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Solution

The correct option is D $\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g)$
The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called standard molar enthalpy of formation.
The option that matches this definition is:
$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g)$

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Similar questions

Q. Choose the reaction(s) from the following options for which the standard enthalpy of reaction is equal to the standard enthalpy of formation:

Standard molar enthalpy of formation, $Δ_{f} H^{⊖}$ is just a special case of enthalpy of reaction, $Δ_{r} H^{⊖}$ . Is the $Δ_{r} H^{⊖}$ for the following reaction same as $Δ_{f} H^{⊖}$ ? Give reason for your answer.
$C a 0 (s) + C O_{2} (g) \to C a C O_{3} (s); Δ_{f} H^{⊖} = - 178.3 k J m o l^{- 1}$