Calculate delta H for this reaction:
N2 + 3H2 --> 2NH3
I am given the bond dissociation energy for N-N (163 kJ/mol), H-H (436 kJ/mol), and N-H (391 kJ/mol)
I can't seem to find a similar problem anywhere, they are all given 2 or 3 similar equations.
thanks!
To calculate the enthalpy change, ΔH, for the given reaction, we can use the bond enthalpy values and apply the bond enthalpy concept.
First, express the reaction in terms of breaking and forming bonds:
N2 + 3H2 → 2NH3
Breaking bonds:
N2 → 2N (1/2 of N-N bond)
3H2 → 6H (3 × 1/2 of H-H bond)
Forming bonds:
6H + 2N → 2NH3 (6 × N-H bond)
Now, let's calculate the bond energy changes:
Energy required to break bonds:
1/2 (N-N) = 1/2 × 163 kJ/mol = 81.5 kJ/mol
3 × 1/2 (H-H) = 3 × 1/2 × 436 kJ/mol = 654 kJ/mol
Energy released when forming bonds:
6 × (N-H) = 6 × 391 kJ/mol = 2346 kJ/mol
Next, sum the bond energy changes for breaking and forming bonds:
(81.5 kJ/mol + 654 kJ/mol) - (2346 kJ/mol) = -1610.5 kJ/mol
The negative sign indicates that the reaction is exothermic, and the magnitude represents the enthalpy change, ΔH, for the reaction. Therefore, the enthalpy change, ΔH, for N2 + 3H2 → 2NH3 is -1610.5 kJ/mol.
To calculate the enthalpy change (∆H) for a reaction, you can use the concept of bond enthalpies. Bond enthalpy refers to the amount of energy required to break a specific bond in a molecule.
In order to calculate the ∆H for the given reaction, you need to consider the bond energies of the bonds being broken and formed. Here's how to calculate it step by step:
1. Determine the bonds that are being broken and formed in the reaction.
In this case:
- N2 has a triple bond between two nitrogen atoms (N≡N).
- H2 has a single bond between two hydrogen atoms (H-H).
- NH3 has three single bonds, one between each nitrogen and hydrogen atom (N-H).
2. Calculate the energy required to break the bonds being broken (endothermic process):
- N2 (triple bond) → 2N: Since there are two nitrogen atoms in N2 and the bond dissociation energy for N-N is given as 163 kJ/mol, we multiply 163 kJ/mol by 2 to break the triple bond. Therefore, the energy required to break the N-N bond is 163 kJ/mol × 2 = 326 kJ/mol.
- 3H2 (single bond) → 6H: Since there are three hydrogen molecules and the bond dissociation energy for H-H is given as 436 kJ/mol, we multiply 436 kJ/mol by 3 to break the single bond in each molecule. Therefore, the energy required to break the H-H bond is 436 kJ/mol × 3 = 1308 kJ/mol.
3. Calculate the energy released when the new bonds are formed (exothermic process):
- 2NH3 (three N-H bonds) → 6H + 2N: The bond dissociation energy for N-H is given as 391 kJ/mol, and we have three N-H bonds in each NH3 molecule. Therefore, the energy released when the new N-H bonds are formed is 391 kJ/mol × 3 = 1173 kJ/mol.
4. Calculate the overall ∆H for the reaction by subtracting the energy used to break the bonds from the energy released in forming the bonds:
∆H = (energy required to break bonds) - (energy released when bonds are formed)
∆H = (326 kJ/mol + 1308 kJ/mol) - 1173 kJ/mol
∆H = 1542 kJ/mol - 1173 kJ/mol
∆H = 369 kJ/mol (rounded to the nearest kJ/mol)
Therefore, the ∆H for the reaction N2 + 3H2 → 2NH3 is 369 kJ/mol.