Calculate the enthalpy change of the reaction | vedclass

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(A) The enthalpy change of a reaction $(\Delta_r H^\circ)$ is calculated using bond energies as follows:$\Delta_r H^\circ = \sum 	ext{Bond energies o

Vedclass · Accepted Answer

$-184 	ext{ kJ mol}^{-1}$

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(A) The enthalpy change of a reaction $(\Delta_r H^\circ)$ is calculated using bond energies as follows:$\Delta_r H^\circ = \sum 	ext{Bond energies of reactants} - \sum 	ext{Bond energies of products}$Step $1$: Calculate the energy required to break the reactant bonds:Energy $= (1 	imes 	ext{Bond energy of } H-H) + (1 	imes 	ext{Bond energy of } Cl-Cl)$Energy $= 436 + 242 = 678 	ext{ kJ mol}^{-1}$Step $2$: Calculate the energy released during the formation of product bonds:Energy $= 2 	imes 	ext{Bond energy of } H-Cl$Energy $= 2 	imes 431 = 862 	ext{ kJ mol}^{-1}$Step $3$: Calculate the enthalpy change:$\Delta_r H^\circ = 678 - 862 = -184 	ext{ kJ mol}^{-1}$

Calculate the enthalpy change of the reaction for $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$, given the bond energies (in $\text{kJ mol}^{-1}$): $H-H = 436$, $Cl-Cl = 242$, $H-Cl = 431$.

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