The enthalpy of formation of ammonia is $- 46.0 K J {m o l}^{- 1}$ . The enthalpy change for the reaction:

$2 {N H}_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$

46.0 K J {m o l}^{- 1}

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92.0 K J {m o l}^{- 1}

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- 23.0 K J {m o l}^{- 1}

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- 92.0 K J {m o l}^{- 1}

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Solution

The correct option is B $92.0 K J {m o l}^{- 1}$
We know that the enthalpy of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Now the enthalpy of formation for ammonia is given to be. $- 46.0 K J m o l^{- 1}$ .for the following equation:
$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g)$ $\to (1)$

Now in the give equation, the number of moles of ammonia is 2 and the reaction in equation (1) is reversed.
Hence, enthalpy of formation of the given reaction is $= 2 \times 46.0 K J m o l^{- 1}$
$= 92 K J m o l^{- 1}$

Hence, option (b) is the correct answer.

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Similar questions

Q. The enthalpy change for a reaction

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

is -92.2 kJ/mol. Calculate the enthalpy of formation of ammonia.

Q. Formation of ammonia is shown by the reaction,

N_{2 (g)} + 3 H_{2 (g)} \to 2 N H_{3 (g)}; △_{r} H^{0}

- 91.8 k J m o l^{- 1}

.
What will be the enthalpy of reaction for the decomposition of

N H_{3}

according to the reaction?

2 N H_{3 (g)} \to N_{2 (g)} + 3 H_{2 (g)}, △_{r} H^{0}

= ?

Q. The enthalpy change for a reaction

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

is -92.2 kJ/mol. Calculate the enthalpy of formation of ammonia.

The value of $Δ_{f} H^{⊖}$ for $N H_{3}$ is -91.8kJ $m o l^{- 1}$ . Calculate enthalpy change for the following reaction.
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$

Q. Bond dissociation enthalpies of

H_{2} (g) a n d N_{2} (g)

are

436.0 k J m o l^{- 1}

and

941.3 k J m o l^{- 1}

respectively. Enthalpy of formation of

N H_{3}

(g) is

- 46.0 k J m o l^{- 1}

. What is the enthalpy of atomization of

N H_{3}

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The enthalpy of formation of ammonia is −46.0KJmol−1. The enthalpy change for the reaction:2NH3(g)→N2(g)+3H2(g)

The enthalpy of formation of ammonia is $- 46.0 K J {m o l}^{- 1}$ . The enthalpy change for the reaction:

$2 {N H}_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$