The enthalpy change for chemical reaction is denoted aas `DeltaH^(Theta)` and `DeltaH^(Theta) = H_(P)^(Theta) - H_(R)^(Theta)`. The relation between enthalpy and internal energy is expressed by equation:
`DeltaH = DeltaU +DeltanRT`
where `DeltaU =` change in internal enegry `Deltan =` change in number of moles, `R =` gas constant.
`H_(2)(g) +((1)/(2))O_(2)(g) = H_(2)O(l), DeltaH_(298K) = - 68.00kcal`
Heat of voporisation of water at `1` atm and `25^(@)C` is `10.00 kcal`. The standard heat of formation (in kcal) of `1` mole vapour a `25^(@)C` is

Options

A. `-78.00`
B. `78.00`
C. `+58.00`
D. `-58.00`

Explanation

To find the standard heat of formation of 1 mole of water vapor at 25°C, we can break the problem down into a series of steps based on the information provided.

### Step 1: Understand the Reaction
The reaction for the formation of water vapor (steam) from its elements is:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2O(g) \]

### Step 2: Identify Given Data
From the question, we have:
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