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Chemistry - Thermochemistry Question with Solution | TestHub

ChemistryThermochemistryEnthalpy of ReactionMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

Anhydrous ammonia is an ultra-clean, energy-dense alternative liquid fuel. It produces no greenhouse gases on combustion. In an experiment, gaseous NH₃ is burned with O₂ in a container of fixed volume according to the equation given below:

$4 N H_{3} (g) + 3 O_{2} (g) ⟶ 2 N_{2} (g) + 6 H_{2} O (l)$

The initial and final states are at $298 K$ . After combustion with $14.40 g$ of O₂, some NH₃ remains unreacted.

Calculate the magnitude of heat energy liberated in KJ . If after combustion with 14.4 gm of $O_{2}$ , some of the $NH_{3}$ remains unreacted at 298 K .

Answer:

27.00

Solution:

$4 NH_{3} (g) + 3 O_{2} (g) ⟶ 2 N_{2} (g) + 6 H_{2} O (ℓ)$ $Δ_{r} H^{\circ} = 6 (- 285.83) - 4 (- 46.11)$

$Δ_{r} H^{\circ} = 1714.98 - 1530.54 = - 184.44$

Energy liberated $= \frac{184.44}{3} \times \frac{14.4}{32} = \frac{184.44}{3} \times 0.45$

$= 61.48 \times 0.45$

$= 27.666$

Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction

⏱ 2mℹ️ Source: QB

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