The value of ∆H (formation) for NH3 is -91.8 kJ per mol than calculate enthalpy change for 2NH3(g)-----> N2(g) + 3H2(g)

18.36 K J

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1836 K J

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183.6 K J

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1.836 K J

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Solution

The correct option is C $183.6 K J$
$Δ H$ for formation is given. For the reverse reaction, $Δ H$ changes sign as the reverse of exothermic reaction will be endothermic. So, $Δ H$ for decomposition is - (-91.8)=91.8 for one mole. But here, two moles are decomposing,
$Δ H = 2 * 91.8 = 183.6 K J$

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Similar questions

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)

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 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:

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

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The value of ∆H (formation) for NH3 is -91.8 kJ per mol than calculate enthalpy change for 2NH3(g)-----> N2(g) + 3H2(g)

The value of ∆H (formation) for NH3 is -91.8 kJ per mol than calculate enthalpy change for 2NH3(g)-----> N2(g) + 3H2(g)

The correct option is C 183.6KJΔH for formation is given. For the reverse reaction,ΔHchanges sign as the reverse of exothermic reaction will be endothermic. So, ΔH for decomposition is - (-91.8)=91.8 for one mole. But here, two moles are decomposing,ΔH=2∗91.8=183.6KJ

The correct option is C $183.6 K J$
$Δ H$ for formation is given. For the reverse reaction, $Δ H$ changes sign as the reverse of exothermic reaction will be endothermic. So, $Δ H$ for decomposition is - (-91.8)=91.8 for one mole. But here, two moles are decomposing,
$Δ H = 2 * 91.8 = 183.6 K J$