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Chemistry - Thermochemistry Question with Solution | TestHub

ChemistryThermochemistryEnthalpy of ReactionEasy2 minPYQ_2020

ChemistryEasysingle choice

The heat of formation of $N H_{3} (g)$ is $- 46 k J m o l^{- 1}$ . The $Δ H$ (in kJ) of the reaction is
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$

Options:

Answer:

Solution:

$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g)$
$Δ H_{r} = - 46 k J / m o l$
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$
$Δ H_{r} = - 2 Δ H_{r}$
$= - 2 (- 46) = 92 k J$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction

⏱ 2mℹ️ Source: PYQ_2020

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