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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Solution

For the reverse reaction, $△$ H changes sign as the reverse of exothermic reaction will be endothermic. So,
$△ H = - (- 45.9) = 45.9$ kj mol $^{- 1}$
But $2$ moles of $N H_{3}$ is present. So,
$△ H = 2 \times 45.9 = 91.8$ kj/mol

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

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

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 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)

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The value of ΔfHo for NH3 is - 45.9, kj mol−1. Calculate enthalpy change for the reaction :2NH3(g)→N2(g)+3H2(g)

For the reverse reaction, △H changes sign as the reverse of exothermic reaction will be endothermic. So,△H=−(−45.9)=45.9 kj mol−1But 2 moles of NH3 is present. So,△H=2×45.9=91.8 kj/mol

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

For the reverse reaction, $△$ H changes sign as the reverse of exothermic reaction will be endothermic. So,
$△ H = - (- 45.9) = 45.9$ kj mol $^{- 1}$
But $2$ moles of $N H_{3}$ is present. So,
$△ H = 2 \times 45.9 = 91.8$ kj/mol