The enthalpy of vaporisation of liquid water using the data
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)` , `DeltaH=-285.77 kJ//mol`
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g)`, `DeltaH=-241.84 kJ//mol`

Options

A. `+43.93`
B. `-43.93`
C. `+527.61`
D. `-527.61`

Answer

To calculate the enthalpy of vaporization of liquid water using the provided data, we will follow these steps:

### Step 1: Write down the two reactions and their enthalpy changes.
1. The formation of liquid water:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H = -285.77 \, \text{kJ/mol} \]

2. The formation of gaseous water:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g) \quad \Delta H = -241.84 \, \text{kJ/mol} \]

### Step 2: Understand the definition of enthalpy of vaporization.
The enthalpy of vaporization (\(\Delta H_{vap}\)) is the amount of energy required to convert a liquid into a gas at constant temperature and pressure. In this case, it can be represented as:
\[ \Delta H_{vap} = \Delta H_{f}(H_2O(g)) - \Delta H_{f}(H_2O(l)) \]

### Step 3: Substitute the values into the equation.
Using the enthalpy values from the reactions:
\[ \Delta H_{vap} = (-241.84 \, \text{kJ/mol}) - (-285.77 \, \text{kJ/mol}) \]

### Step 4: Simplify the equation.
This simplifies to:
\[ \Delta H_{vap} = -241.84 \, \text{kJ/mol} + 285.77 \, \text{kJ/mol} \]

### Step 5: Calculate the final value.
Now, performing the calculation:
\[ \Delta H_{vap} = 285.77 - 241.84 = 43.93 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of vaporization of liquid water is:
\[ \Delta H_{vap} = +43.93 \, \text{kJ/mol} \]

---