What is the bond enthalpy of N2(g)+2H2(g) --> N2H4(g)?
To determine the bond enthalpy of a reaction, we need to know the bond enthalpies of the bonds broken and formed in the reaction. The bond enthalpy is the average amount of energy required to break a particular type of bond in a molecule in the gas phase.
Let's break down the reaction, N2(g) + 2H2(g) --> N2H4(g), and determine the bonds involved:
1. In N2(g) and H2(g), both molecules are diatomic, meaning they consist of two atoms bonded together. For N2, there is a triple bond between two nitrogen atoms, and for H2, there is a single bond between two hydrogen atoms.
2. In N2H4(g), the molecule consists of two nitrogen atoms and four hydrogen atoms. There are four N-H bonds in N2H4.
Now, let's calculate the bond enthalpy of each bond using average bond enthalpy values:
- The average bond enthalpy of the N-N bond in N2 is approximately 945 kJ/mol.
- The average bond enthalpy of the H-H bond in H2 is approximately 436 kJ/mol.
- The average bond enthalpy of the N-H bond is approximately 391 kJ/mol.
Based on the stoichiometry of the reaction, we can determine the total bond enthalpy change as follows:
∆H = (2 * ∆H(N-H)) - (∆H(N-N) + 2 * ∆H(H-H))
Substituting the values:
∆H = (2 * 391 kJ/mol) - (945 kJ/mol + 2 * 436 kJ/mol)
Simplifying:
∆H = 782 kJ/mol - 945 kJ/mol - 872 kJ/mol
∆H = -1035 kJ/mol
Therefore, the bond enthalpy change (∆H) for the reaction N2(g) + 2H2(g) --> N2H4(g) is -1035 kJ/mol. Note that the negative sign indicates an exothermic reaction, meaning energy is released in the form of heat.
The bond enthalpy is the energy required to break a bond in a molecule. To calculate the bond enthalpy of a reaction, you need to know the individual bond enthalpies of the bonds in the reactants and products.
For the reaction N2(g) + 2H2(g) -> N2H4(g), you need to determine the bond enthalpies of the respective bonds: N≡N, H-H, and N-H.
The bond enthalpy of N≡N is 941 kJ/mol.
The bond enthalpy of H-H is 436 kJ/mol.
The bond enthalpy of N-H is typically around 391 kJ/mol.
Since there are two N-H bonds formed and one N≡N bond and two H-H bonds broken, you calculate the total change in bond enthalpy as follows:
[(2 * N-H bond enthalpies formed) + (1 * N≡N bond enthalpy broken)] - (2 * H-H bond enthalpies broken)
[(2 * 391 kJ/mol) + (1 * 941 kJ/mol)] - (2 * 436 kJ/mol)
= 782 kJ/mol + 941 kJ/mol - 872 kJ/mol
= 851 kJ/mol
Therefore, the bond enthalpy of the reaction N2(g) + 2H2(g) -> N2H4(g) is 851 kJ/mol.