The enthalpy of formation of ammonia is $- 46.0 k J m o l^{- 1}$ . The enthalpy change for the reaction.
$2 N H_{3} (g) \to 2 N_{2} (g) + 3 H_{2} (g)$ is:

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Solution

Given, Heat of formation $= - 46.0 K J / m o l e$
Since, the enthalpy of formation of ammonia is the enthalpy change for the formation of $1$ mole of ammonia from its element.
$2 N H_{3} (g) \to 2 N_{2} (g) + 3 H_{2} (g)$
$△ H_{r e a c t i o n} = 2 \times △ H_{f o r m a t i o n} = 2 \times - 46.0 K J / m o l e$
$H e a t o f r e a c t i o n = - 92.0 K J / m o l e$

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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. 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}

= ?

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

Given, Heat of formation =−46.0 KJ/moleSince, the enthalpy of formation of ammonia is the enthalpy change for the formation of 1 mole of ammonia from its element.2NH3(g)→2N2(g)+3H2(g)△Hreaction=2×△Hformation=2×−46.0 KJ/moleHeat of reaction=−92.0 KJ/mole

The enthalpy of formation of ammonia is $- 46.0 k J m o l^{- 1}$ . The enthalpy change for the reaction.
$2 N H_{3} (g) \to 2 N_{2} (g) + 3 H_{2} (g)$ is: