Enthalpy change for the reaction, 4 H(g) 2 H _ 2 (g) is - 869.6 kJ The dissociation energy of H-H bond is

Correct answer: B. +434.8 ~kJ array ll 4 H(g) & arrow 2 H_2(g) ; Δ H=-869.6 ~kJ 2 H_2(g) & arrow 4 H(g) ; Δ H=869.6 ~kJ H_2(g) & arrow 2 H(g) ; Δ H=(869.6)/(2)=434.8 ~kJ array

NEET 2011

Exam: NEET
Year: 2011
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2011 · Chemistry · Thermodynamics (C)

Enthalpy change for the reaction, 4 H(g) 2 H _ 2 (g) is - 869.6 kJ...

Enthalpy change for the reaction, 4

\mathrm{H}(g) \longrightarrow 2 \mathrm{H}

}_{2}(g)

\text{ is }-

869.6

\mathrm{ kJ}

The dissociation energy of

\mathrm{H}-\mathrm{H}

bond is

Options

A $-869.6 \mathrm{~kJ}$
B $+434.8 \mathrm{~kJ}$
C $+217.4 \mathrm{~kJ}$
D $-434.8 \mathrm{~kJ}$

Correct answer

B. $+434.8 \mathrm{~kJ}$

Step-by-step solution

\begin{array}{ll} 4 \mathrm{H}(g) & \rightarrow 2 \mathrm{H}_2(g) ; \Delta H=-869.6 \mathrm{~kJ} \\ 2 \mathrm{H}_2(g) & \rightarrow 4 \mathrm{H}(g) ; \Delta H=869.6 \mathrm{~kJ} \\ \mathrm{H}_2(g) & \rightarrow 2 \mathrm{H}(g) ; \Delta H=\frac{869.6}{2}=434.8 \mathrm{~kJ} \end{array}

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