In order to decompose 9 grams of water 142.5 KJ heat is required. Hence the enthalpy of formation of water is

Options

A. `+285 kJ`
B. `-285 kJ`
C. `+142.5 kJ`
D. `-142.5 kJ`

Answer

To find the enthalpy of formation of water based on the given information, we can follow these steps:

### Step 1: Understand the decomposition of water
The decomposition of water can be represented by the reaction:
\[ \text{H}_2\text{O} \rightarrow \text{H}_2 + \frac{1}{2} \text{O}_2 \]
The enthalpy change (\( \Delta H \)) for this reaction is given as +142.5 kJ for 9 grams of water.

### Step 2: Calculate the molar mass of water
The molar mass of water (H₂O) is:
\[ \text{Molar mass of H}_2\text{O} = 2 \times 1 + 16 = 18 \text{ g/mol} \]

### Step 3: Determine the amount of heat for 1 mole of water
Since 9 grams of water require 142.5 kJ for decomposition, we can find the heat required for 18 grams (1 mole) of water:
\[ \text{Heat for 18 g} = \left( \frac{142.5 \text{ kJ}}{9 \text{ g}} \right) \times 18 \text{ g} = 285 \text{ kJ} \]

### Step 4: Relate the decomposition to formation
The enthalpy of formation of water is the reverse of the decomposition reaction. Therefore, the enthalpy of formation (\( \Delta H_f \)) can be expressed as:
\[ \Delta H_f = -285 \text{ kJ/mol} \]

### Conclusion
Thus, the enthalpy of formation of water is:
\[ \Delta H_f = -285 \text{ kJ/mol} \]

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