The bond dissociation enthalpies of H_2, Cl_2 | vedclass

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(A) The chemical equation for the formation of $1 	ext{ mole}$ of $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) ightarrow HCl(g)$.Enthalpy of

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$-93 	ext{ kJ mol}^{-1}$

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(A) The chemical equation for the formation of $1 	ext{ mole}$ of $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) ightarrow HCl(g)$.Enthalpy of formation $(\Delta_f H^\circ)$ is calculated using bond enthalpies: $\Delta_f H^\circ = \sum 	ext{Bond Enthalpies of Reactants} - \sum 	ext{Bond Enthalpies of Products}$.$\Delta_f H^\circ = [\frac{1}{2} \Delta H_{H-H} + \frac{1}{2} \Delta H_{Cl-Cl}] - [\Delta H_{H-Cl}]$.Substituting the given values: $\Delta_f H^\circ = [\frac{1}{2}(434) + \frac{1}{2}(242)] - 431$.$\Delta_f H^\circ = [217 + 121] - 431$.$\Delta_f H^\circ = 338 - 431 = -93 	ext{ kJ mol}^{-1}$.

The bond dissociation enthalpies of $H_2$, $Cl_2$, and $HCl$ are $434 \text{ kJ mol}^{-1}$, $242 \text{ kJ mol}^{-1}$, and $431 \text{ kJ mol}^{-1}$ respectively. Calculate the enthalpy of formation of $HCl$.

Similar Questions

Calculate the enthalpy change for the following reaction, using the given bond energies $(\text{kJ/mol})$: $C-H = 414$, $H-O = 463$, $H-Cl = 431$, $C-Cl = 326$, and $C-O = 335$.
$CH_3OH(g) + HCl(g) \rightarrow CH_3Cl(g) + H_2O(g)$

Calculate $\Delta H$ for the reaction: $H_{2(g)} + O_{2(g)} \rightarrow H_2O_{2(g)}$ given the bond energies: $BE_{H-H} = 436 \ kJ/mol$,$BE_{O=O} = 499 \ kJ/mol$,$BE_{O-O} = 142 \ kJ/mol$,and $BE_{O-H} = 460 \ kJ/mol$. (in $kJ$)

Given that $H_{2(g)} + 1/2 O_{2(g)} \to H_2O_{(g)}; \Delta H_1$ and $H_{2(g)} + 1/2 O_{2(g)} \to H_2O_{(l)}; \Delta H_2$,which of the following is correct?

The standard heats of formation for $CH_4$, $C_2H_4$, and $C_3H_8$ are $-17.9$, $12.5$, and $-24.8 \ kcal/mol$ respectively. The heat of reaction $(\Delta H)$ for the reaction $CH_4 + C_2H_4 \rightarrow C_3H_8$ in $kcal$ is:

Explain the standard enthalpy of formation.

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