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:

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 $92.0 k J m o l^{- 1}$

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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 2 N_{2} (g) + 3 H_{2} (g)

is:

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 standard enthalpy of formation of $N H_{3}$ is $46.0 k J m o l^{- 1}$ . If the enthalpy of formation of $H_{2}$ from its atom is $- 436 k J m o l^{- 1}$ and that of $N_{2}$ is $- 712 k J m o l^{- 1}$ , the average bond enthalpy of $N - H$ bond in $N H_{3}$ is

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

The correct option is B 92.0 kJ mol−1

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:

The correct option is B $92.0 k J m o l^{- 1}$