Answer :
To calculate the enthalpy of formation of ammonium perchlorate ([tex]\(NH_4ClO_4\)[/tex]), we can use the given reaction and the enthalpy change of the reaction. Here's a step-by-step solution:
1. Understand the Reaction:
The thermochemical equation provided is:
[tex]\[
2 \, NH_4ClO_4(s) \rightarrow N_2(g) + Cl_2(g) + 2 \, O_2(g) + 4 \, H_2O(g)
\][/tex]
2. Given Data:
- The enthalpy change of the reaction ([tex]\(\Delta H_{reaction}\)[/tex]) is [tex]\(-375.6 \, \text{kJ}\)[/tex].
- The enthalpy of formation of water ([tex]\(H_2O(g)\)[/tex]) is [tex]\(-241.8 \, \text{kJ/mol}\)[/tex].
3. Identify the Products' Enthalpies:
As we produce 4 moles of water, we can calculate the total enthalpy contributed by the formation of water vapor:
[tex]\[
\Delta H_{water} = 4 \, \text{moles} \times (-241.8 \, \text{kJ/mol}) = -967.2 \, \text{kJ}
\][/tex]
4. Relate the Reaction Enthalpy to Formation Enthalpies:
The enthalpy change for the decomposition can be expressed as:
[tex]\[
\Delta H_{reaction} = [\text{Enthalpies of formation of products}] - [\text{Enthalpies of formation of reactants}]
\][/tex]
5. Calculate the Enthalpy of Formation for Ammonium Perchlorate:
Rearrange the enthalpy equation to solve for the enthalpy of formation of ammonium perchlorate. For one mole of [tex]\(NH_4ClO_4\)[/tex], since the reaction involves 2 moles, we adjust accordingly:
[tex]\[
[-967.2 \, \text{kJ} \text{ (Enthalpy for water)}] - [2 \times \text{Enthalpy of } NH_4ClO_4] = -375.6 \, \text{kJ}
\][/tex]
Solving for the enthalpy of formation of [tex]\(NH_4ClO_4\)[/tex]:
[tex]\[
2 \times [\text{Enthalpy of } NH_4ClO_4] = \Delta H_{water} + \Delta H_{reaction}
\][/tex]
[tex]\[
2 \times [\text{Enthalpy of } NH_4ClO_4] = -967.2 + 375.6 = -591.6 \, \text{kJ}
\][/tex]
6. Find the Final Value:
[tex]\[
[\text{Enthalpy of } NH_4ClO_4] = \frac{-591.6}{2} = -295.8 \, \text{kJ/mol}
\][/tex]
Therefore, the enthalpy of formation for ammonium perchlorate is [tex]\(-295.8 \, \text{kJ/mol}\)[/tex]. The correct answer is C. [tex]\(-295.8 \, \text{kJ}\)[/tex].
1. Understand the Reaction:
The thermochemical equation provided is:
[tex]\[
2 \, NH_4ClO_4(s) \rightarrow N_2(g) + Cl_2(g) + 2 \, O_2(g) + 4 \, H_2O(g)
\][/tex]
2. Given Data:
- The enthalpy change of the reaction ([tex]\(\Delta H_{reaction}\)[/tex]) is [tex]\(-375.6 \, \text{kJ}\)[/tex].
- The enthalpy of formation of water ([tex]\(H_2O(g)\)[/tex]) is [tex]\(-241.8 \, \text{kJ/mol}\)[/tex].
3. Identify the Products' Enthalpies:
As we produce 4 moles of water, we can calculate the total enthalpy contributed by the formation of water vapor:
[tex]\[
\Delta H_{water} = 4 \, \text{moles} \times (-241.8 \, \text{kJ/mol}) = -967.2 \, \text{kJ}
\][/tex]
4. Relate the Reaction Enthalpy to Formation Enthalpies:
The enthalpy change for the decomposition can be expressed as:
[tex]\[
\Delta H_{reaction} = [\text{Enthalpies of formation of products}] - [\text{Enthalpies of formation of reactants}]
\][/tex]
5. Calculate the Enthalpy of Formation for Ammonium Perchlorate:
Rearrange the enthalpy equation to solve for the enthalpy of formation of ammonium perchlorate. For one mole of [tex]\(NH_4ClO_4\)[/tex], since the reaction involves 2 moles, we adjust accordingly:
[tex]\[
[-967.2 \, \text{kJ} \text{ (Enthalpy for water)}] - [2 \times \text{Enthalpy of } NH_4ClO_4] = -375.6 \, \text{kJ}
\][/tex]
Solving for the enthalpy of formation of [tex]\(NH_4ClO_4\)[/tex]:
[tex]\[
2 \times [\text{Enthalpy of } NH_4ClO_4] = \Delta H_{water} + \Delta H_{reaction}
\][/tex]
[tex]\[
2 \times [\text{Enthalpy of } NH_4ClO_4] = -967.2 + 375.6 = -591.6 \, \text{kJ}
\][/tex]
6. Find the Final Value:
[tex]\[
[\text{Enthalpy of } NH_4ClO_4] = \frac{-591.6}{2} = -295.8 \, \text{kJ/mol}
\][/tex]
Therefore, the enthalpy of formation for ammonium perchlorate is [tex]\(-295.8 \, \text{kJ/mol}\)[/tex]. The correct answer is C. [tex]\(-295.8 \, \text{kJ}\)[/tex].
