To determine the enthalpy change of the reaction ([tex]\( \Delta H_{\text{rxn}} \)[/tex]) for the dissociation of NH[tex]\(_4\)[/tex]Br in water, follow these steps:
### Step 1: Find the Mass of the Solution
The total mass of the solution is the sum of the mass of NH[tex]\(_4\)[/tex]Br and the mass of water.
- Mass of NH[tex]\(_4\)[/tex]Br: [tex]\( 4.0 \, \text{g} \)[/tex]
- Mass of water: [tex]\( 46.0 \, \text{g} \)[/tex]
[tex]\[ \text{Mass of solution} = 4.0 \, \text{g} + 46.0 \, \text{g} = 50.0 \, \text{g} \][/tex]
### Step 2: Calculate the Change in Temperature ([tex]\( \Delta T \)[/tex])
The change in temperature is the final temperature minus the initial temperature.
- Initial temperature: [tex]\( 21.3^\circ \text{C} \)[/tex]
- Final temperature: [tex]\( 18.0^\circ \text{C} \)[/tex]
[tex]\[ \Delta T = 18.0^\circ \text{C} - 21.3^\circ \text{C} = -3.3^\circ \text{C} \][/tex]
### Step 3: Calculate the Heat Absorbed/Released by the Solution
The heat ([tex]\( q \)[/tex]) absorbed or released by the solution is given by:
[tex]\[ q_{\text{solution}} = \text{mass of solution} \times \text{specific heat of the solution} \times \Delta T \][/tex]
- Specific heat of solution ([tex]\( C_{\text{soln}} \)[/tex]): [tex]\( 4.18 \, \text{J/g}^\circ \text{C} \)[/tex]
- Mass of solution: [tex]\( 50.0 \, \text{g} \)[/tex]
- [tex]\( \Delta T \)[/tex]: [tex]\( -3.3^\circ \text{C} \)[/tex]
[tex]\[ q_{\text{solution}} = 50.0 \, \text{g} \times 4.18 \, \text{J/g}^\circ \text{C} \times (-3.3^\circ \text{C}) = -689.7 \, \text{J} \][/tex]
### Step 4: Calculate the Heat Absorbed/Released by the Calorimeter
The heat ([tex]\( q \)[/tex]) absorbed or released by the calorimeter is given by:
[tex]\[ q_{\text{calorimeter}} = \text{heat capacity of calorimeter} \times \Delta T \][/tex]
- Heat capacity of calorimeter ([tex]\( C_{\text{cal}} \)[/tex]): [tex]\( 5.7 \, \text{J/}^\circ \text{C} \)[/tex]
- [tex]\( \Delta T \)[/tex]: [tex]\( -3.3^\circ \text{C} \)[/tex]
[tex]\[ q_{\text{calorimeter}} = 5.7 \, \text{J/}^\circ \text{C} \times (-3.3^\circ \text{C}) = -18.81 \, \text{J} \][/tex]
### Step 5: Calculate the Total Heat Absorbed/Released
The total heat ([tex]\( q_{\text{total}} \)[/tex]) absorbed or released by the system is the sum of the heat by the solution and the calorimeter.
[tex]\[ q_{\text{total}} = q_{\text{solution}} + q_{\text{calorimeter}} \][/tex]
[tex]\[ q_{\text{total}} = -689.7 \, \text{J} + (-18.81 \, \text{J}) = -708.51 \, \text{J} \][/tex]
Convert [tex]\( q_{\text{total}} \)[/tex] from J to kJ.
[tex]\[ q_{\text{total}} = -708.51 \, \text{J} \times \left( \frac{1 \, \text{kJ}}{1000 \, \text{J}} \right) = -0.70851 \, \text{kJ} \][/tex]
### Step 6: Calculate the Moles of NH[tex]\(_4\)[/tex]Br
First, find the molar mass of NH[tex]\(_4\)[/tex]Br.
- Molar mass of NH[tex]\(_4\)[/tex]Br: [tex]\( 97.94 \, \text{g/mol} \)[/tex]
Determine the moles of NH[tex]\(_4\)[/tex]Br used:
[tex]\[ \text{Moles of NH}_4\text{Br} = \frac{\text{Mass of NH}_4\text{Br}}{\text{Molar Mass of NH}_4\text{Br}} \][/tex]
[tex]\[ \text{Moles of NH}_4\text{Br} = \frac{4.0 \, \text{g}}{97.94 \, \text{g/mol}} = 0.040841331 \, \text{mol} \][/tex]
### Step 7: Determine the Enthalpy Change ([tex]\( \Delta H_{\text{rxn}} \)[/tex])
Finally, calculate the enthalpy change per mole of NH[tex]\(_4\)[/tex]Br.
[tex]\[ \Delta H_{\text{rxn}} = \frac{q_{\text{total}}}{\text{moles of NH}_4\text{Br}} \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = \frac{-0.70851 \, \text{kJ}}{0.040841331 \, \text{mol}} = -17.34786735 \, \text{kJ/mol} \][/tex]
### Final Answer
[tex]\[ \Delta H_{\text{rxn}} = -17.35 \, \text{kJ/mol} \][/tex]
The enthalpy change of the reaction is [tex]\( -17.35 \, \text{kJ/mol} \)[/tex]. This indicates the process is exothermic, as indicated by the negative sign.
