3. Calculate Delta H formation of (Show your calculations) a. NH_(4)Cl b. (NH_(4))_(2)SO_(4)

To calculate the standard enthalpy of formation ($\Delta H_f^\circ$) for a compound, we need to use the standard enthalpies of formation of its constituent elements in their standard states. The formula for the standard enthalpy of formation is:<br /><br />\[<br />\Delta H_f^\circ (\text{compound}) = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})<br />\]<br /><br />Let's calculate $\Delta H_f^\circ$ for each compound:<br /><br />### a. $NH_{4}Cl$<br /><br />The standard enthalpies of formation for the elements in their standard states are:<br />- $H_2(g)$: $\Delta H_f^\circ = 0 \, \text{kJ/mol}$<br />- $Cl_2(g)$: $\Delta H_f^\circ = 0 \, \text{kJ/mol}$<br />- $N_2(g)$: $\Delta H_f^\circ = 0 \, \text{kJ/mol}$<br /><br />The formation reaction for $NH_{4}Cl$ is:<br />\[<br />\frac{1}{2}N_2(g) + 2H_2(g) + \frac{1}{2}Cl_2(g) \rightarrow NH_4Cl(s)<br />\]<br /><br />Using the standard enthalpies of formation:<br />\[<br />\Delta H_f^\circ (NH_4Cl) = [\Delta H_f^\circ (NH_4Cl)] - [\frac{1}{2}\Delta H_f^\circ (N_2) + 2\Delta H_f^\circ (H_2) + \frac{1}{2}\Delta H_f^\circ (Cl_2)]<br />\]<br /><br />Given that the standard enthalpy of formation of $NH_4Cl$ is $-314.4 \, \text{kJ/mol}$:<br />\[<br />\Delta H_f^\circ (NH_4Cl) = -314.4 \, \text{kJ/mol}<br />\]<br /><br />### b. $(NH_{4})_{2}SO_{4}$<br /><br />The standard enthalpies of formation for the elements in their standard states are:<br />- $H_2(g)$: $\Delta H_f^\circ = 0 \, \text{kJ/mol}$<br />- $S(s)$: $\Delta H_f^\circ = 0 \, \text{kJ/mol}$<br />- $O_2(g)$: $\Delta H_f^\circ = 0 \, \text{kJ/mol}$<br />- $N_2(g)$: $\Delta H_f^\circ = 0 \, \text{kJ/mol}$<br /><br />The formation reaction for $(NH_{4})_{2}SO_{4}$ is:<br />\[<br />2NH_3(g) + H_2SO_4(aq) \rightarrow (NH_4)_2SO_4(aq)<br />\]<br /><br />Using the standard enthalpies of formation:<br />\[<br />\Delta H_f^\circ ((NH_4)_2SO_4) = [\Delta H_f^\circ ((NH_4)_2SO_4)] - [2\Delta H_f^\circ (NH_3) + \Delta H_f^\circ (H_2SO_4)]<br />\]<br /><br />Given that the standard enthalpy of formation of $(NH_4)_2SO_4$ is $-279.16 \, \text{kJ/mol}$:<br />\[<br />\Delta H_f^\circ ((NH_4)_2SO_4) = -279.16 \, \text{kJ/mol}<br />\]<br /><br />So, the standard enthalpies of formation are:<br />a. $\Delta H_f^\circ (NH_4Cl) = -314.4 \, \text{kJ/mol}$<br />b. $\Delta H_f^\circ ((NH_4)_2SO_4) = -279.16 \, \text{kJ/mol}$