Answer :
To determine the change in enthalpy for the reaction given:
[tex]\[ N_2 + 3 \, H_2 \rightarrow 2 \, NH_3 \][/tex]
we need to follow these steps:
### Step 1: Bonds Broken in Reactants
First, we identify the bonds broken in the reactants:
- There is 1 triple bond in [tex]\( N_2 \)[/tex] (Nitrogen molecule).
- There are 3 single bonds in [tex]\( H_2 \)[/tex] (Hydrogen molecules).
Using the bond energies provided, we can calculate the total energy required to break these bonds.
- The bond energy for [tex]\( N≡N \)[/tex] is 942 kJ/mol.
- The bond energy for [tex]\( H-H \)[/tex] is 432 kJ/mol.
Thus, the energy required to break these bonds is:
[tex]\[ \text{Energy to break } N_2 = 1 \times 942 \, \text{kJ} = 942 \, \text{kJ} \][/tex]
[tex]\[ \text{Energy to break } 3 \, H_2 = 3 \times 432 \, \text{kJ} = 1296 \, \text{kJ} \][/tex]
The total energy required to break the bonds in the reactants is:
[tex]\[ \text{Total energy broken} = 942 \, \text{kJ} + 1296 \, \text{kJ} = 2238 \, \text{kJ} \][/tex]
### Step 2: Bonds Formed in Products
Next, we identify the bonds formed in the products:
- In 2 molecules of [tex]\( NH_3 \)[/tex], there are 6 N-H bonds (since each [tex]\( NH_3 \)[/tex] molecule has 3 N-H bonds).
Using the bond energies provided, we can calculate the total energy released in forming these bonds.
- The bond energy for [tex]\( N-H \)[/tex] is 386 kJ/mol.
Thus, the energy released in forming these bonds is:
[tex]\[ \text{Energy to form } 6 \, N-H = 6 \times 386 \, \text{kJ} = 2316 \, \text{kJ} \][/tex]
### Step 3: Calculate the Change in Enthalpy (ΔH)
The change in enthalpy (ΔH) for the reaction is calculated from the energy required to break the bonds minus the energy released in forming the bonds.
[tex]\[ \Delta H = \text{Total energy broken} - \text{Total energy formed} \][/tex]
[tex]\[ \Delta H = 2238 \, \text{kJ} - 2316 \, \text{kJ} \][/tex]
[tex]\[ \Delta H = -78 \, \text{kJ} \][/tex]
### Final Answer
Expressed to two significant figures, the enthalpy change for the reaction is:
[tex]\[ \boxed{-78 \, \text{kJ}} \][/tex]
[tex]\[ N_2 + 3 \, H_2 \rightarrow 2 \, NH_3 \][/tex]
we need to follow these steps:
### Step 1: Bonds Broken in Reactants
First, we identify the bonds broken in the reactants:
- There is 1 triple bond in [tex]\( N_2 \)[/tex] (Nitrogen molecule).
- There are 3 single bonds in [tex]\( H_2 \)[/tex] (Hydrogen molecules).
Using the bond energies provided, we can calculate the total energy required to break these bonds.
- The bond energy for [tex]\( N≡N \)[/tex] is 942 kJ/mol.
- The bond energy for [tex]\( H-H \)[/tex] is 432 kJ/mol.
Thus, the energy required to break these bonds is:
[tex]\[ \text{Energy to break } N_2 = 1 \times 942 \, \text{kJ} = 942 \, \text{kJ} \][/tex]
[tex]\[ \text{Energy to break } 3 \, H_2 = 3 \times 432 \, \text{kJ} = 1296 \, \text{kJ} \][/tex]
The total energy required to break the bonds in the reactants is:
[tex]\[ \text{Total energy broken} = 942 \, \text{kJ} + 1296 \, \text{kJ} = 2238 \, \text{kJ} \][/tex]
### Step 2: Bonds Formed in Products
Next, we identify the bonds formed in the products:
- In 2 molecules of [tex]\( NH_3 \)[/tex], there are 6 N-H bonds (since each [tex]\( NH_3 \)[/tex] molecule has 3 N-H bonds).
Using the bond energies provided, we can calculate the total energy released in forming these bonds.
- The bond energy for [tex]\( N-H \)[/tex] is 386 kJ/mol.
Thus, the energy released in forming these bonds is:
[tex]\[ \text{Energy to form } 6 \, N-H = 6 \times 386 \, \text{kJ} = 2316 \, \text{kJ} \][/tex]
### Step 3: Calculate the Change in Enthalpy (ΔH)
The change in enthalpy (ΔH) for the reaction is calculated from the energy required to break the bonds minus the energy released in forming the bonds.
[tex]\[ \Delta H = \text{Total energy broken} - \text{Total energy formed} \][/tex]
[tex]\[ \Delta H = 2238 \, \text{kJ} - 2316 \, \text{kJ} \][/tex]
[tex]\[ \Delta H = -78 \, \text{kJ} \][/tex]
### Final Answer
Expressed to two significant figures, the enthalpy change for the reaction is:
[tex]\[ \boxed{-78 \, \text{kJ}} \][/tex]
