Question

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

• A. - 434.8 kJ
• B. - 869.6 kJ
• C. #ERROR!
• D. #ERROR!

Answer 1

The Correct Option is C

To find the dissociation energy of the $H-H$ bond, let's analyze the given reaction and enthalpy change:

The reaction given is:

$4H_{(g)} \rightarrow 2H_{2(g)}$

The enthalpy change for this reaction is -869.6 \, \text{kJ}. This reaction represents the combination of atomic hydrogen to form $H_2$ molecules. Since the enthalpy change is negative, it's an exothermic process.

In order to determine the dissociation energy of a single $H-H$ bond, we can reverse the given reaction to represent the breaking of $H_2$ into hydrogen atoms:

$2H_{2(g)} \rightarrow 4H_{(g)}$

Reversing the reaction will change the sign of the enthalpy change:

\Delta H = +869.6 \, \text{kJ}

This is the energy required for the dissociation of 2 moles of $H_2$ molecules into 4 moles of hydrogen atoms. Therefore, the dissociation energy for one mole of $H_2$ (which contains two $H-H$ bonds) is:

\text{Dissociation energy of } H_2 = \frac{869.6}{2} \, \text{kJ/mol} = 434.8 \, \text{kJ/mol}

Hence, the dissociation energy of one $H-H$ bond is:

434.8 \, \text{kJ/mol}

Thus, the correct option is:

Option: 434.8 kJ

The initially provided correct answer #ERROR! seems incorrect based on this analysis.