To determine the change in enthalpy (ΔH) for the reaction [tex]\( N_2 + 3 H_2 \rightarrow 2 NH_3 \)[/tex], we need to compare the energy required to break bonds in the reactants to the energy released when new bonds form in the products.
First, let's consider the bonds in the reactants:
- In [tex]\( N_2 \)[/tex], there is one triple bond, [tex]\( N \equiv N \)[/tex], which has a bond energy of 942 kJ/mol.
- In [tex]\( 3 H_2 \)[/tex], there are three single bonds, [tex]\( H - H \)[/tex], each with a bond energy of 432 kJ/mol.
Energy required to break the bonds in the reactants:
- Breaking 1 bond of [tex]\( N \equiv N \)[/tex] requires [tex]\( 1 \times 942 \)[/tex] kJ.
- Breaking 3 bonds of [tex]\( H - H \)[/tex] requires [tex]\( 3 \times 432 \)[/tex] kJ.
So, the total energy required to break the bonds is:
[tex]\[ 942 \text{ kJ} + 3 \times 432 \text{ kJ} = 942 \text{ kJ} + 1296 \text{ kJ} = 2238 \text{ kJ} \][/tex]
Next, let’s consider the bonds in the products:
- In [tex]\( 2 NH_3 \)[/tex], each [tex]\( NH_3 \)[/tex] molecule has three [tex]\( N - H \)[/tex] bonds. Therefore, two [tex]\( NH_3 \)[/tex] molecules have six [tex]\( N - H \)[/tex] bonds.
- Each [tex]\( N - H \)[/tex] bond has a bond energy of 386 kJ/mol.
Energy released to form the bonds in the products:
- Forming 6 bonds of [tex]\( N - H \)[/tex] releases [tex]\( 6 \times 386 \)[/tex] kJ.
So, the total energy released when the bonds are formed is:
[tex]\[ 6 \times 386 \text{ kJ} = 2316 \text{ kJ} \][/tex]
Finally, the change in enthalpy (ΔH) for the reaction is the energy released minus the energy required:
[tex]\[ \Delta H = \text{Energy released} - \text{Energy required} = 2316 \text{ kJ} - 2238 \text{ kJ} = 78 \text{ kJ} \][/tex]
Therefore, the enthalpy change for the reaction is [tex]\( 78 \)[/tex] kilojoules, expressed to two significant figures.
