Question

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

Answer 1

Answer:

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

Explanation:

The standard enthalpy of the reaction ($\Delta H_{r}^{\circ}$) 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.

$\Delta H_{r}^{\circ} = \sum \Delta H_{f}^{\circ} (Products) - \sum \Delta H_{f}^{\circ} (reactants)$

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

Therefore,

a. H₂O (l) + 1/2 O₂ (g) → H₂O₂(l)

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

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

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

b. N₂ (g) + O₂ (g) → 2NO (g)

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

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

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

c. 2H₂ (g) + O₂ (g) → 2H₂O (g)

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

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

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

d. 2H₂ (g) + O₂ (g) → 2H₂O (l)

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

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

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

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

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

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

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