Question

Calculate the standard enthalpy of change for each of the following reactions:
CaO(s) + 2 HCl(g) --> CaCl2(s) + H2O(g)
N2(g) + 4 H2O (g) --> N2O4 (g) + 4 H2(g

Answer 1

The standard enthalpy of change can be calculated using the Hess's law

$\Delta H_{reaction} = \sum_i\Delta H_{f, products} - \sum_i\Delta H_{f, reactants}$

and the standard enthalpy of formation for each of the compounds:

$\Delta H_f (\text{CaO})=-635$ kJ/mol

$\Delta H_f (\text{HCl})=−92.31$ kJ/mol

$\Delta H_f (\text{CaCl}_2)=−795.42$ kJ/mol

$\Delta H_f (\text{H}_2\text{O, g})=−241.818$ kJ/mol

$\Delta H_f (\text{N}_2\text{O}_4)=+9.16$ kJ/mol.

All elements in their standard states have a standard enthalpy of formation of zero, therefore the standard enthalpy of formation of H₂(g) and N₂(g) are equal to zero.

Thus, the standard enthalpy change of the reaction CaO(s) + 2 HCl(g) $\rightarrow$ CaCl₂(s) + H₂O(g) is:

$\Delta H = \Delta H_f (\text{H}_2\text{O}) + \Delta H_f (\text{CaCl}_2) - 2·\Delta H_f (\text{HCl}) - \Delta H_f (\text{CaO})$

$\Delta H =-241.818 -795.42-2·(-92.31) +635 = -217.618$ kJ/mol

and the standard enthalpy change of the reaction N₂(g) + 4H₂O (g) $\rightarrow$ N₂O₄ (g) + 4H₂(g) is:

$\Delta H = \Delta H_f (\text{N}_2\text{O}_4) - 4· \Delta H_f (\text{H}_2\text{O})$

$\Delta H = 9.16-4·(-241.818) = 976.432$ kJ/mol.

Answer: the standard change of enthalpy for the following reactions:

CaO(s) + 2 HCl(g) $\rightarrow$ CaCl₂(s) + H₂O(g) is -217.618 kJ/mol

N₂(g) + 4 H₂O (g) --> N₂O₄ (g) + 4 H₂(g) is 976.432 kJ/mol.

Therefore, the first reaction is exothermic and the second reaction is endothermic.