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

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

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

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

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+ 45.9 k J m o l^{- 1}

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Solution

The correct option is B $+ 91.8 k J m o l^{- 1}$
The given reaction is:-
$N_{2} (g) + 3 H_{2} (g) \to 2 {N H}_{3} (g); △ r H^{0} = - 91.8 K J / m o l$
Nor, for the reaction
$2 {N H}_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$
$△ r H^{0} = - (- 91.8) = + 91.8 K J / m o l$
(since this reaction is reverse reaction of the above given reaction so the sign of $△ H$ will get reversed).

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Similar questions

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. 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)

The enthalpy of formation of $N H_{3}$ (g) is $- 46.2$ kJ mol $^{- 1}$ . The heat of the following reaction is:

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

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g) : △_{r} H^{⊖} = - 92.4 k J m o l^{- 1}

What is the standard enthalpy of formation of

N H_{3}

gas?

Q. The value of

Δ f H^{o}

for

N H_{3}

is - 45.9, kj

m o l^{- 1}

. Calculate enthalpy change for the reaction :

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

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Formation of ammonia is shown by the reaction, N2(g)+3H2(g)→2NH3(g);△rH0 = −91.8kJmol−1.What will be the enthalpy of reaction for the decomposition of NH3 according to the reaction?2NH3(g)→N2(g)+3H2(g),△rH0 = ?

The correct option is B +91.8kJmol−1The given reaction is:- N2(g)+3H2(g)→2NH3(g);△rH0=−91.8KJ/molNor, for the reaction 2NH3(g)→N2(g)+3H2(g) △rH0=−(−91.8)=+91.8KJ/mol(since this reaction is reverse reaction of the above given reaction so the sign of △H will get reversed).

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