SATHEE: Chemical Thermodynamics Question 487

Chemical Thermodynamics Question 487

Question: Calculate the heat change in the reaction $4, N H_{3} (g) + 3 O_{2} (g) \to 2 N_{2} (g) + 6 H_{2} O (ℓ)$ at 298 K, given that the heats of formation at 298 K for $N H_{3} (g)$ and $H_{2} O (ℓ)$ are - 46.0 and - 28.60 kJ $m o l^{- 1}$ respectively.

Options:

A) - 1861 kJ

B) - 1361 kJ

C) - 1261 kJ

D) - 1532 kJ

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Answer:

Correct Answer: D

Solution:

$Δ H^{o}$ for the reaction $4 N H_{3} (g) + 3 O_{2} (g) \to 6 H_{2} O (l) + 2 N_{2} (g)$

$Δ H^{o} = Δ H_{f}^{o} (p r o d u c t s) - Δ H_{f}^{o} (r e a c t a n t s)$

$= 6 H_{f}^{o} [H_{2} O (l)] + Δ {H^{\circ}}_{f} [N_{2} (g)]$

$- 4 Δ H [N H_{3} (g)] + 3 Δ H_{f} [O_{2} (g)]$

$Δ H_{f}^{o} [H_{2} O (l)] = - 286.0 k J m o l^{- 1}$

$Δ H_{f}^{o} [N H_{2} (g)] = 0$ and $Δ H_{f}^{o} [N_{2} (g)]$

= 0 (by convention) $Δ H^{o} = 6 (- 286) + 2 (0) - 4 (- 46.0) + 3 (0)$

$= - 1716 + 184 = - 1532, k J$

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Chemical Thermodynamics Question 487

Question: Calculate the heat change in the reaction 4,NH3(g)+3O2(g)→2N2(g)+6H2O(ℓ) at 298 K, given that the heats of formation at 298 K for NH3(g) and H2O(ℓ) are - 46.0 and - 28.60 kJ mol−1 respectively.

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Question: Calculate the heat change in the reaction $4, N H_{3} (g) + 3 O_{2} (g) \to 2 N_{2} (g) + 6 H_{2} O (ℓ)$ at 298 K, given that the heats of formation at 298 K for $N H_{3} (g)$ and $H_{2} O (ℓ)$ are - 46.0 and - 28.60 kJ $m o l^{- 1}$ respectively.