Respuesta :
Answer:
The correct option is: D) C (graphite)+ 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l)
Explanation:
The standard enthalpy change of a given chemical reaction ([tex]\Delta H^{\circ }_{r}[/tex]) is equal to the difference of the sum of formation enthalpy ([tex]\Delta H^{\circ }_{f}[/tex]) of products and the sum of formation enthalpy of reactants.
[tex]\Delta H^{\circ }_{r} = \sum v. \Delta H^{\circ }_{f}(products) - \sum v. \Delta H^{\circ }_{f}(reactants)[/tex]
The standard enthalpy of formation is zero for elements that exist in their standard states.
The standard enthalpy change of a reaction is equal to the standard enthalpy of formation of methanol (l) [[tex]\Delta H^{\circ }_{r} = \Delta H^{\circ }_{f}(CH_{3}OH, l)[/tex]], only if:
1. The standard enthalpy of all the other reactants and products should be zero, i.e. all the other reactants and products should be in their standard states.
2. The stoichiometric coefficient of methanol (CH₃OH, l) should be 1.
Among the given options, only the option D satisfies both the conditions.
D) C (graphite)+ 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l)
The standard enthalpy change for this chemical reaction:
[tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ] - \left [\Delta H^{\circ }_{f}(C, s, graphite) + \Delta H^{\circ }_{f}(H_{2}, g) + \Delta H^{\circ }_{f}(O_{2}, g)\right ][/tex]
The standard enthalpy of formation:
[tex]\Delta H^{\circ }_{f}(C, s, graphite) = 0 kJ/mol[/tex]
[tex]\Delta H^{\circ }_{f}(H_{2}, g) = 0 kJ/mol[/tex]
[tex]\Delta H^{\circ }_{f}(O_{2}, g) = 0 kJ/mol[/tex]
⇒ [tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ] - \left [0 kJ/mol + 0 kJ/mol + 0 kJ/mol\right ] [/tex]
⇒ [tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ][/tex]
Therefore, the correct option is (D).
