Hydrazine has the formula N2H4 and is used as a rocket fuel (e.g. for the Apollo moon rockets). It burns in the following reaction for which the enthalpy change is -583 kJ/mol.

N2H4(g) + O2(g) → N2(g) + 2 H2O(g)
Calculate the N-N bond enthalpy in hydrazine given the following bond enthalpies.
Bond enthalpies: N-H 388, O=O 498, NòN 944, O-H 463 kJ/mo

To calculate the N-N bond enthalpy in hydrazine, we need to use the bond enthalpies given in the question. The N-N bond enthalpy is the energy required to break the N-N bond in hydrazine.

Using the given bond enthalpies:
N-H = 388 kJ/mol
O=O = 498 kJ/mol
N≡N = 944 kJ/mol
O-H = 463 kJ/mol

First, let's look at the balanced chemical equation:
N2H4(g) + O2(g) → N2(g) + 2H2O(g)

In this equation, the N-N bond is broken in N2H4 and formed in N2.

The bond broken is N≡N, and there are 2 of these bonds in N2H4, so the total energy required to break them is:
2 x (N≡N) = 2 x 944 kJ/mol = 1888 kJ/mol

Next, let's look at the bond formed. In the product N2, there is only 1 N-N bond.

The bond formed is N≡N, so the energy released in forming the N-N bond is:
1 x (N≡N) = 1 x 944 kJ/mol = 944 kJ/mol

To find the N-N bond enthalpy in hydrazine, we subtract the energy released from the energy required to break the bond:
N-N bond enthalpy = Energy required to break the bond - Energy released in forming the bond
N-N bond enthalpy = 1888 kJ/mol - 944 kJ/mol = 944 kJ/mol

Therefore, the N-N bond enthalpy in hydrazine is 944 kJ/mol.

To calculate the N-N (nitrogen-nitrogen) bond enthalpy in hydrazine (N2H4), we need to calculate the energy required to break the N-N bond in N2H4.

The given reaction is:
N2H4(g) + O2(g) → N2(g) + 2H2O(g)

From the balanced reaction, we can see that 1 mole of N2H4 produces 1 mole of N2. Therefore, the enthalpy change for breaking the N-N bond can be calculated by comparing the enthalpy change of the given reaction with the enthalpies of formation of the other compounds involved.

The enthalpy change for the given reaction is -583 kJ/mol.

Based on the bond enthalpies given, we can calculate the enthalpy change for breaking the N-N bond in N2H4.

Step 1: Calculate the energy required to break the N-H bonds in N2H4
There are 4 N-H bonds in N2H4, so the energy required to break these bonds is 4 * (388 kJ/mol) = 1552 kJ/mol.

Step 2: Calculate the energy required to break the O=O bond in O2
There is 1 O=O bond in O2, so the energy required to break this bond is 1 * (498 kJ/mol) = 498 kJ/mol.

Step 3: Calculate the energy released in the formation of N2 gas and H2O gas
1 mole of N2 is formed, so the energy released is 1 * (0 kJ/mol) = 0 kJ/mol.
2 moles of H2O are formed, so the energy released is 2 * (0 kJ/mol) = 0 kJ/mol.

Step 4: Calculate the total energy change for the reaction involving the N-N bond
The total energy change for the reaction can be calculated by subtracting the energy required to break the bonds from the energy released in the formation of the products.
Total energy change = -583 kJ/mol - (1552 kJ/mol + 498 kJ/mol + 0 kJ/mol + 0 kJ/mol)
Total energy change = -583 kJ/mol - 2050 kJ/mol
Total energy change = -2633 kJ/mol

Step 5: Calculate the N-N bond enthalpy in hydrazine
Since 1 mole of N-N bond is broken, the N-N bond enthalpy can be calculated by dividing the total energy change by the number of moles of N-N bond.
N-N bond enthalpy = (-2633 kJ/mol)/1 = -2633 kJ/mol

Therefore, the N-N bond enthalpy in hydrazine is approximately -2633 kJ/mol.

148