Answer :
To determine the enthalpy change for the given reaction, we need to use the formula:
[tex]\[ \Delta H_{reaction} = \sum \left(\Delta H_{f, \text{products}}\right) - \sum \left(\Delta H_{f, \text{reactants}}\right) \][/tex]
First, let us write down the enthalpies of formation for the substances involved in the reaction:
- Enthalpy of formation of [tex]\( NH_3 \)[/tex] [tex]\( (\Delta H_f(NH_3)) = -45.9 \, \text{kJ/mol} \)[/tex]
- Enthalpy of formation of [tex]\( H_2O \)[/tex] [tex]\( (\Delta H_f(H_2O)) = -241.8 \, \text{kJ/mol} \)[/tex]
- Enthalpy of formation of [tex]\( NO \)[/tex] [tex]\( (\Delta H_f(NO)) = 91.3 \, \text{kJ/mol} \)[/tex]
- Enthalpy of formation of [tex]\( O_2 \)[/tex] [tex]\( (\Delta H_f(O_2)) = 0 \, \text{kJ/mol} \)[/tex] (since [tex]\( O_2 \)[/tex] is a diatomic element in its standard state)
The given reaction is:
[tex]\[ 4 \, NH_3(g) + 5 \, O_2(g) \rightarrow 6 \, H_2O(g) + 4 \, NO(g) \][/tex]
Next, we calculate the sum of the enthalpies of formation for the products:
[tex]\[ \sum \left( \Delta H_{f, \text{products}} \right) = \left( 6 \times \Delta H_f(H_2O) \right) + \left( 4 \times \Delta H_f(NO) \right) \][/tex]
Substituting the values:
[tex]\[ \sum \left( \Delta H_{f, \text{products}} \right) = \left( 6 \times -241.8 \right) + \left( 4 \times 91.3 \right) \][/tex]
[tex]\[ \sum \left( \Delta H_{f, \text{products}} \right) = -1450.8 + 365.2 = -1085.6 \, \text{kJ} \][/tex]
Then, we calculate the sum of the enthalpies of formation for the reactants:
[tex]\[ \sum \left( \Delta H_{f, \text{reactants}} \right) = \left( 4 \times \Delta H_f(NH_3) \right) + \left( 5 \times \Delta H_f(O_2) \right) \][/tex]
Substituting the values:
[tex]\[ \sum \left( \Delta H_{f, \text{reactants}} \right) = \left( 4 \times -45.9 \right) + \left( 5 \times 0 \right) \][/tex]
[tex]\[ \sum \left( \Delta H_{f, \text{reactants}} \right) = -183.6 \, \text{kJ} \][/tex]
Finally, we find the enthalpy change of the reaction:
[tex]\[ \Delta H_{reaction} = \sum \left( \Delta H_{f, \text{products}} \right) - \sum \left( \Delta H_{f, \text{reactants}} \right) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{reaction} = -1085.6 - (-183.6) \][/tex]
[tex]\[ \Delta H_{reaction} = -1085.6 + 183.6 = -902 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the reaction is [tex]\(-902 \, \text{kJ}\)[/tex]. The correct answer is:
[tex]\[ \boxed{-902 \, \text{kJ}} \][/tex]
[tex]\[ \Delta H_{reaction} = \sum \left(\Delta H_{f, \text{products}}\right) - \sum \left(\Delta H_{f, \text{reactants}}\right) \][/tex]
First, let us write down the enthalpies of formation for the substances involved in the reaction:
- Enthalpy of formation of [tex]\( NH_3 \)[/tex] [tex]\( (\Delta H_f(NH_3)) = -45.9 \, \text{kJ/mol} \)[/tex]
- Enthalpy of formation of [tex]\( H_2O \)[/tex] [tex]\( (\Delta H_f(H_2O)) = -241.8 \, \text{kJ/mol} \)[/tex]
- Enthalpy of formation of [tex]\( NO \)[/tex] [tex]\( (\Delta H_f(NO)) = 91.3 \, \text{kJ/mol} \)[/tex]
- Enthalpy of formation of [tex]\( O_2 \)[/tex] [tex]\( (\Delta H_f(O_2)) = 0 \, \text{kJ/mol} \)[/tex] (since [tex]\( O_2 \)[/tex] is a diatomic element in its standard state)
The given reaction is:
[tex]\[ 4 \, NH_3(g) + 5 \, O_2(g) \rightarrow 6 \, H_2O(g) + 4 \, NO(g) \][/tex]
Next, we calculate the sum of the enthalpies of formation for the products:
[tex]\[ \sum \left( \Delta H_{f, \text{products}} \right) = \left( 6 \times \Delta H_f(H_2O) \right) + \left( 4 \times \Delta H_f(NO) \right) \][/tex]
Substituting the values:
[tex]\[ \sum \left( \Delta H_{f, \text{products}} \right) = \left( 6 \times -241.8 \right) + \left( 4 \times 91.3 \right) \][/tex]
[tex]\[ \sum \left( \Delta H_{f, \text{products}} \right) = -1450.8 + 365.2 = -1085.6 \, \text{kJ} \][/tex]
Then, we calculate the sum of the enthalpies of formation for the reactants:
[tex]\[ \sum \left( \Delta H_{f, \text{reactants}} \right) = \left( 4 \times \Delta H_f(NH_3) \right) + \left( 5 \times \Delta H_f(O_2) \right) \][/tex]
Substituting the values:
[tex]\[ \sum \left( \Delta H_{f, \text{reactants}} \right) = \left( 4 \times -45.9 \right) + \left( 5 \times 0 \right) \][/tex]
[tex]\[ \sum \left( \Delta H_{f, \text{reactants}} \right) = -183.6 \, \text{kJ} \][/tex]
Finally, we find the enthalpy change of the reaction:
[tex]\[ \Delta H_{reaction} = \sum \left( \Delta H_{f, \text{products}} \right) - \sum \left( \Delta H_{f, \text{reactants}} \right) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{reaction} = -1085.6 - (-183.6) \][/tex]
[tex]\[ \Delta H_{reaction} = -1085.6 + 183.6 = -902 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the reaction is [tex]\(-902 \, \text{kJ}\)[/tex]. The correct answer is:
[tex]\[ \boxed{-902 \, \text{kJ}} \][/tex]
