contestada

For the following reactions, the ΔH° rxn is not equal to ΔH° f for the product except for ________.

a. H2O (l) + 1/2 O2 (g) → H2O2(l)
b. N2 (g) + O2 (g) → 2NO (g)
c. 2H2 (g) + O2 (g) → 2H2O (g)
d. 2H2 (g) + O2 (g) → 2H2O (l)
e. 2C(s, graphite) + 2H2(g) → C2H4 (g)

Respuesta :

Answer:

The correct option is: e. 2C(s, graphite) + 2H₂(g) → C₂H₄ (g)

Explanation:

The standard enthalpy of the reaction ([tex]\Delta H_{r}^{\circ}[/tex]) is the change in enthalpy associated with a given chemical reaction. It can be calculated by the standard enthalpies of formation of the products and reactants.

[tex]\Delta H_{r}^{\circ} = \sum \Delta H_{f}^{\circ} (Products) - \sum \Delta H_{f}^{\circ} (reactants)[/tex]

The standard enthalpy of formation ([tex]\Delta H_{f}^{\circ}[/tex]) of an elements that is present in its standard state is zero.

Therefore,

a. H₂O (l) + 1/2 O₂ (g) → H₂O₂(l)

[tex]\Delta H_{r}^{\circ} = \left [\Delta H_{f}^{\circ} (H_{2}O_{2}, l) \right ] - \left [\Delta H_{f}^{\circ} (H_{2}O, l)+ 1/2\times \Delta H_{f}^{\circ} (O_{2}, g)\right][/tex]

Here, the [tex]\Delta H_{f}^{\circ} (O_{2}, g) = 0 kJ/mol[/tex]

→ [tex]\Delta H_{r}^{\circ} = \left [\Delta H_{f}^{\circ} (H_{2}O_{2}, l) \right ] - \left [\Delta H_{f}^{\circ} (H_{2}O, l)\right] \neq \Delta H_{f}^{\circ} (Products)[/tex]

b. N₂ (g) + O₂ (g) → 2NO (g)

[tex]\Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (NO, g) \right ] - \left [\Delta H_{f}^{\circ} (N_{2}, g)+\Delta H_{f}^{\circ} (O_{2}, g)\right][/tex]

Here, the [tex]\Delta H_{f}^{\circ} (O_{2}, g) = \Delta H_{f}^{\circ} (N_{2}, g)= 0 kJ/mol[/tex]

→ [tex]\Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (NO, g) \right ] \neq \Delta H_{f}^{\circ} (Product, NO)[/tex]

c. 2H₂ (g) + O₂ (g) → 2H₂O (g)

[tex]\Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (H_{2}O, g) \right ] - \left [2\times \Delta H_{f}^{\circ} (H_{2}, g)+\Delta H_{f}^{\circ} (O_{2}, g)\right][/tex]

Here, the [tex]\Delta H_{f}^{\circ} (O_{2}, g) = \Delta H_{f}^{\circ} (H_{2}, g)= 0 kJ/mol[/tex]

→ [tex]\Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (H_{2}O, g) \right ] \neq \Delta H_{f}^{\circ} (Product, H_{2}O)[/tex]

d. 2H₂ (g) + O₂ (g) → 2H₂O (l)

[tex]\Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (H_{2}O, l) \right ] - \left [2\times \Delta H_{f}^{\circ} (H_{2}, g)+\Delta H_{f}^{\circ} (O_{2}, g)\right][/tex]

Here, the [tex]\Delta H_{f}^{\circ} (O_{2}, g) = \Delta H_{f}^{\circ} (H_{2}, g)= 0 kJ/mol[/tex]

→ [tex]\Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (H_{2}O, l) \right ] \neq \Delta H_{f}^{\circ} (Product, H_{2}O)[/tex]

e. 2C(s, graphite) + 2H₂(g) → C₂H₄ (g)

[tex]\Delta H_{r}^{\circ} = \left [\Delta H_{f}^{\circ} (C_{2}H_{4}, g) \right ] - \left [2\times \Delta H_{f}^{\circ} (C, s graphite)+2\times \Delta H_{f}^{\circ} (H_{2}, g)\right][/tex]

Here, the [tex]\Delta H_{f}^{\circ} (C, s, graphite) = \Delta H_{f}^{\circ} (H_{2}, g)= 0 kJ/mol[/tex]

→ [tex]\Delta H_{r}^{\circ} = \left [\Delta H_{f}^{\circ} (C_{2}H_{4}, g) \right ] = \Delta H_{f}^{\circ} (Product, C_{2}H_{4})[/tex]

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