If the bond energies of H-H, Br-Br , and H-Br are 433,192 and 364 ~kJ /mol^ -1 respectively, the Δ H^ for the reaction H_ 2(g) +Br_ 2(g) arrow 2 HBr_ (g) is:

Correct answer: D. -103 ~kJ H-H+Br-Br arrow 2 H-Br array rrr 433+192 & 2 & ×364 =625 & =728 array Energy absorbed Energy released Net energy released = H_ R - H_ p =625-728=-103 ~kJ . Related Theory Fluorine has high bond energy because it is smallest among these molecules so it is tightly bonded to other atoms due to greater nuclear charge.

NEET 2004

Exam: NEET
Year: 2004
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2004 · Chemistry · Thermodynamics (C)

If the bond energies of H-H, Br-Br , and H-Br are 433,192 and 364...

If the bond energies of

\mathrm{H}-\mathrm{H}, \mathrm{Br}-\mathrm{Br}

, and

\mathrm{H}-\mathrm{Br}

are 433,192 and

364 \mathrm{~kJ} /\mathrm{mol}^{-1}

respectively, the

\Delta \mathrm{H}^{\circ}

for the reaction

\mathrm{H}_{2(g)}+\mathrm{Br}_{2(g)} \rightarrow 2 \mathrm{HBr}_{(g)}

is:

Options

A $-261 \mathrm{~kJ}$
B $+103 \mathrm{~kJ}$
C $+261 \mathrm{~kJ}$
D $-103 \mathrm{~kJ}$

Correct answer

D. $-103 \mathrm{~kJ}$

Step-by-step solution

\mathrm{H}-\mathrm{H}+\mathrm{Br}-\mathrm{Br} \rightarrow 2 \mathrm{H}-\mathrm{Br}

\begin{array}{rrr}433+192 & 2 & \times364 \\ =625 & =728 \end{array}

Energy absorbed Energy released Net energy released

=\Sigma \mathrm{H}_{\mathrm{R}}-\Sigma \mathrm{H}_{\mathrm{p}}=625-728=-103 \mathrm{~kJ}$. Related Theory Fluorine has high bond energy because it is smallest among these molecules so it is tightly bonded to other atoms due to greater nuclear charge.

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