Question

write the equation for bond enthalpy of Cl-Cl bond in the Cl2 molecule.∆fh⁰ for dissociation of Cl2 molecule is 242.7 kJ
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Answer 1

Answer:

The equation for the bond enthalpy of the Cl-Cl bond in the $Cl_2$ molecule is

$Cl_2_{(g)}\ --->\ Cl_{(g)}\ +\ Cl_{(g)}$·

Explanation:

Given that,

The $\Delta_{f}H^{0}$ for the dissociation of the $Cl_2$ molecule  $=$  $+\ 242.7$ KJ

So, the given reaction is the dissociation of the $Cl_2$ molecule·

The equation for the dissociation of the  $Cl_2$ follows,

  $Cl_2_{(g)}\ --->\ Cl_{(g)}\ +\ Cl_{(g)}$

Here,

The Chlorine molecule dissociates forms two chloride gases· Heat is required when the Cl-Cl bond is broken and heat is released when a new bond forms·

The formula for the standard formation enthalpy follows,

  $\Delta_{f}H^{0} \ \ =\ \ \Sigma \Delta_{f}H^{0}(products)\ -\ \Sigma \Delta_{f}H^{0}(reactants)$

Here, the formation enthalpy of the product is greater than the formation enthalpy of the reactants· Which is why the value is positive·

Therefore,

The equation for the bond enthalpy of the Cl-Cl bond in the $Cl_2$ molecule is

$Cl_2_{(g)}\ --->\ Cl_{(g)}\ +\ Cl_{(g)}$·