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

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 $92.0 K J m o l^{- 1}$
The enthalpy of formation of ammonia is $- 46.0 k J m o l^{- 1}$ .
$1 / 2 N_{2} (g) + 3 / 2 H_{2} (g) \to N H_{3} (g) Δ H = - 46.0 k J / m o l$

Multiply above equation with 2.
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g) Δ H = - 46.0 \times 2 = - 92.0 k J / m o l$

Reverse above equation
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g) Δ H = + 92.0 k J / m o l$

The enthalpy change for the reaction $2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$ is $92.0 K J m o l^{- 1}$

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Similar questions

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)

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}

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.
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}

= ?

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

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

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