Question

For which one of the following equations is ∆Hº_react equal to ∆Hº_f for the product:

• A. \(N_2(g)+O_3(g)→N_2O_3(g)\)
• B. \(CH_4(g)+2Cl_2(g)→CH_2Cl_2(l)+2HCl(g)\)
• C. \(Xe(g)+2F_2(g)→XeF_4(g)\)
• D. \(2CO(g)+O_2(g)→2CO_2(g)\)

Answer 1

The Correct Option is C

The given question asks us to identify which of the listed equations has the standard enthalpy change of reaction (\(\Delta H^\circ_{\text{react}}\)) equal to the standard enthalpy of formation (\(\Delta H^\circ_f\)) for the product. This condition is true when the product is formed from its elements in their standard states.

Let's examine each option to determine if \(\Delta H^\circ_{\text{react}} = \Delta H^\circ_f\).

1. \(N_2(g) + O_3(g) \rightarrow N_2O_3(g)\)
   In this reaction, the compound \(N_2O_3\) is not formed from elements in their standard states. \(O_3\) is not the most stable form of oxygen; \(O_2\) is. Thus, \(\Delta H^\circ_{\text{react}}\) is not equal to \(\Delta H^\circ_f\).
2. \(CH_4(g) + 2Cl_2(g) \rightarrow CH_2Cl_2(l) + 2HCl(g)\)
   Here, \(CH_2Cl_2\) is not being formed from its elements in their most stable forms; methane and chlorine gas are not in their elemental forms required for the formation of \(CH_2Cl_2\). Therefore, \(\Delta H^\circ_{\text{react}}\) is not equal to \(\Delta H^\circ_f\).
3. \(Xe(g) + 2F_2(g) \rightarrow XeF_4(g)\)
   The product \(XeF_4\) is formed from \(Xe\) and \(F_2\), the elemental forms of xenon and fluorine, respectively. This reaction satisfies the condition required for \(\Delta H^\circ_{\text{react}} = \Delta H^\circ_f\).
4. \(2CO(g) + O_2(g) \rightarrow 2CO_2(g)\)
   Even though \(CO_2\) is being formed, the reaction involves \(CO\) as a reactant, which is not an elemental form. Therefore, \(\Delta H^\circ_{\text{react}}\) is not equal to the \(\Delta H^\circ_f\) as the products are not directly formed from elemental carbon and oxygen.

Given the analysis of each option, the correct answer is \(Xe(g) + 2F_2(g) \rightarrow XeF_4(g)\) where the reaction fulfills the criteria for \(\Delta H^\circ_{\text{react}} = \Delta H^\circ_f\).