2. Calculate (Show your calculations) a. Delta H_(neut) for NH_(4)Cl b Delta H_(neut)for(NH_(4))_(2)SO_(4) c. Delta H_(diss) for NH_(4)Cl d. Delta H_(diss) for (NH_(4))_(2)SO_(4)

To calculate the enthalpy changes for the given compounds, we need to use the concept of standard enthalpy of formation ($\Delta H_f^\circ$) and the formula for standard enthalpy of reaction ($\Delta H_{rxn}^\circ$).<br /><br />### a. $\Delta H_{neut}$ for $NH_{4}Cl$<br /><br />The neutralization reaction for $NH_{4}Cl$ can be written as:<br />\[ NH_{4}Cl (aq) + H_{2}O (l) \rightarrow NH_{3} (aq) + HCl (aq) \]<br /><br />However, $NH_{4}Cl$ is a salt and does not neutralize with water in the same way acids and bases do. Instead, it dissociates in water. Therefore, the concept of neutralization does not apply here.<br /><br />### b. $\Delta H_{neut}$ for $(NH_{4})_{2}SO_{4}$<br /><br />Similarly, $(NH_{4})_{2}SO_{4}$ is an ammonium salt and does not undergo a neutralization reaction with water. It dissociates in water, so neutralization does not apply.<br /><br />### c. $\Delta H_{diss}$ for $NH_{4}Cl$<br /><br />The dissociation of $NH_{4}Cl$ in water can be represented as:<br />\[ NH_{4}Cl (aq) \rightarrow NH_{4}^+ (aq) + Cl^- (aq) \]<br /><br />The standard enthalpy of dissociation ($\Delta H_{diss}$) can be calculated using the standard enthalpy of formation values:<br />\[ \Delta H_{diss} = \Delta H_f^\circ (NH_{4}^+) + \Delta H_f^\circ (Cl^-) - \Delta H_f^\circ (NH_{4}Cl) \]<br /><br />Using standard enthalpy of formation values:<br />- $\Delta H_f^\circ (NH_{4}Cl (s)) = -314.4 \, \text{kJ/mol}$<br />- $\Delta H_f^\circ (NH_{4}^+ (aq)) = -66.4 \, \text{kJ/mol}$<br />- $\Delta H_f^\circ (Cl^- (aq)) = -167.2 \, \text{kJ/mol}$<br /><br />\[ \Delta H_{diss} = (-66.4) + (-167.2) - (-314.4) \]<br />\[ \Delta H_{diss} = -233.6 \, \text{kJ/mol} \]<br /><br />### d. $\Delta H_{diss}$ for $(NH_{4})_{2}SO_{4}$<br /><br />The dissociation of $(NH_{4})_{2}SO_{4}$ in water can be represented as:<br />\[ (NH_{4})_{2}SO_{4} (aq) \rightarrow 2 NH_{4}^+ (aq) + SO_{4}^{2-} (aq) \]<br /><br />The standard enthalpy of dissociation ($\Delta H_{diss}$) can be calculated using the standard enthalpy of formation values:<br />\[ \Delta H_{diss} = 2 \cdot \Delta H_f^\circ (NH_{4}^+) + \Delta H_f^\circ (SO_{4}^{2-}) - \Delta H_f^\circ ((NH_{4})_{2}SO_{4}) \]<br /><br />Using standard enthalpy of formation values:<br />- $\Delta H_f^\circ ((NH_{4})_{2}SO_{4} (s)) = -305.6 \, \text{kJ/mol}$<br />- $\Delta H_f^\circ (NH_{4}^+ (aq)) = -66.4 \, \text{kJ/mol}$<br />- $\Delta H_f^\circ (SO_{4}^{2-} (aq)) = -300.2 \, \text{kJ/mol}$<br /><br />\[ \Delta H_{diss} = 2 \cdot (-66.4) + (-300.2) - (-305.6) \]<br />\[ \Delta H_{diss} = -132.8 - 300.2 + 305.6 \]<br />\[ \Delta H_{diss} = -127.4 \, \text{kJ/mol} \]<br /><br />### Summary<br /><br />- **a.** $\Delta H_{neut}$ for $NH_{4}Cl$: Not applicable as it does not neutralize with water.<br />- **b.** $\Delta H_{neut}$ for $(NH_{4})_{2}SO_{4}$: Not applicable as it does not neutralize with water.<br />- **c.** $\Delta H_{diss}$ for $NH_{4}Cl$: $-233.6 \, \text{kJ/mol}$<br />- **d.** $\Delta H_{diss}$ for $(NH