SATHEE: Thermodynamics Question 95

Thermodynamics Question 95

Question: Given the bond energies $N \equiv N, H - H$ and $N - H$ bonds are $945, 436$ and $391 k J m o l e^{- 1}$ respectively, the enthalpy of the following reaction $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ is

[EAMCET 1992; JIPMER 1997]

Options:

A) $- 93 k J$

B) $102 k J$

C) $90 k J$

D) $105 k J$

Show Answer

Answer:

Correct Answer: A

Solution:

$\underset{\begin{matrix} Energy absorbed \end{matrix}}{\underset{945 + 3 \times 436}{N \equiv N + 3 H - H}} \overset{}{\to} \underset{\begin{matrix} Energy released \end{matrix}}{\underset{2 \times (3 \times 391) = 2346}{\overset{H}{\overset{|}{\underset{H}{\underset{|}{2 N}} -}} H}}$ Net. energy released = 2346 - 2253 = 93 kJ i.e. ΔH = - 93 kJ .

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Thermodynamics Question 95

Question: Given the bond energies N≡N,H−H and N−H bonds are 945,436 and 391kJmole−1 respectively, the enthalpy of the following reaction N2(g)+3H2(g)→2NH3(g) is

Options:

Answer:

Solution:

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Question: Given the bond energies $N \equiv N, H - H$ and $N - H$ bonds are $945, 436$ and $391 k J m o l e^{- 1}$ respectively, the enthalpy of the following reaction $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ is