Hydrazine is used as a rocket fuel because its reaction with oxygen is extremely exothermic:

N2H4(liquid) + O2(g)  N2(g) + 2H2O(liquid) ΔH(reaction) = -615 kJ mol-1
What is the enthalpy of this reaction if water were produced in the gaseous rather than the liquid form? Note that 40.7 kJ mol-1 is the molar enthalpy of vaporization of water.
A) -534 kJ mol-1
B) -656 kJ mol-1
C) -696 kJ mol-1
D) -574 kJ mol-1
E) -615 kJ mol-1

Some of the heat generated would have been diverted to convert 2H2O(l) to 2H2O(g),

so heat generated would be reduced by 2*40.7=81.4 kJ/mol.
Can you take it from here?

E) -615 kJ mol-1

To find the enthalpy of the reaction if water were produced in the gaseous form, we need to consider the enthalpy change for the vaporization of water.

The balanced equation for the reaction is:
N2H4(liquid) + O2(g) → N2(g) + 2H2O(liquid)

Given: ΔH(reaction) = -615 kJ/mol
ΔH(vaporization) = 40.7 kJ/mol

Since 2 mol of water are produced in the reaction, we need to consider the enthalpy change for the vaporization of 2 mol of water.

The enthalpy change for vaporization of 2 mol of water is:
ΔH(vaporization) × 2 = 40.7 kJ/mol × 2 = 81.4 kJ/mol

To find the enthalpy of the reaction if water were produced in the gaseous form, we subtract the enthalpy change for vaporization from the overall enthalpy change of the reaction:

ΔH(reaction) - ΔH(vaporization) = -615 kJ/mol - 81.4 kJ/mol = -696.4 kJ/mol

The enthalpy of the reaction if water were produced in the gaseous form is -696.4 kJ/mol.

Therefore, the correct answer is C) -696 kJ/mol.

To determine the enthalpy of the reaction if water were produced in the gaseous form, we need to take into account the molar enthalpy of vaporization of water. The molar enthalpy of vaporization represents the amount of energy required to convert 1 mole of liquid water into vapor at a given temperature.

In this case, we have:

N2H4(liquid) + O2(g) -> N2(g) + 2H2O(liquid)

To produce gaseous water, we need to convert the liquid water into vapor. Since we have 2 moles of liquid water being produced in the reaction, we will need to consider the molar enthalpy of vaporization of water twice.

The given enthalpy change of the reaction is -615 kJ mol^-1, which means that 1 mole of N2H4 reacts to produce -615 kJ of energy.

The molar enthalpy of vaporization of water is given as 40.7 kJ mol^-1. Therefore, to convert 2 moles of liquid water into vapor, we would need 2 * 40.7 kJ of energy.

Now, we can calculate the enthalpy change with water in gaseous form:

Enthalpy change = -615 kJ + 2 * 40.7 kJ
Enthalpy change = -615 kJ + 81.4 kJ
Enthalpy change = -533.6 kJ

Therefore, the enthalpy of the reaction if water were produced in gaseous form is approximately -533.6 kJ mol^-1.

Comparing the calculated value with the given options, the closest value is option A) -534 kJ mol^-1.