Calculate the enthalpy of vaporisation for water form the following:
`H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(g),DeltaH =- 57.0 kcal`
`H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l),DeltaH =- 68.3 kcal`
Also calculate the heat required to change `1g H_(2)O(l)` to `H_(2)O(g)`

Explanation

To calculate the enthalpy of vaporization for water and the heat required to change 1 g of H₂O(l) to H₂O(g), we can follow these steps:

### Step 1: Write down the given reactions and their enthalpy changes.
1. \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g), \Delta H = -57.0 \, \text{kcal} \) (Equation 1)
2. \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l), \Delta H = -68.3 \, \text{kcal} \) (Equation 2)

### Step 2: Rearrange the equations to find the enthalpy of vaporization.
To find the enthalpy of vaporization (\( \Delta H_{vap} \)), we need to convert \( H_2O(l) \) to \( H_2O(g) \). We can do this by subtracting Equation 2 from Equation 1:
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