Calorific value of `H_2` is `-143KJ g^(-1)` . Hence, `Delta H_(f)^(@)` of `H_2O` is:

Options

A. `-143 kJ mol^(-1)`
B. `-286 kJ mol^(-1)`
C. `+143 kJ mol^(-1)`
D. `+286 kJ mol^(-1)`

Explanation

To find the enthalpy of formation (ΔH_f) of water (H₂O) from the given calorific value of hydrogen (H₂), we can follow these steps:

### Step 1: Understand the calorific value
The calorific value of hydrogen is given as -143 kJ/kg. This means that when 1 kg of hydrogen is combusted, it releases 143 kJ of energy.

### Step 2: Convert calorific value to per gram
To find the energy released per gram of hydrogen, we can convert the calorific value from kg to g:
\[ \text{Calorific value per gram} = \frac{-143 \text{ kJ}}{1000 \text{ g}} = -0.143 \text{ kJ/g} \]

### Step 3: Relate the combustion of hydrogen to the formation of water
The balanced equation for the combustion of hydrogen is:
\[ 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2O(l) \]
From this equation, we see that 2 moles of hydrogen produce 2 moles of water.

### Step 4: Calculate the energy released for 2 grams of hydrogen
Since the molecular weight of H₂ is 2 g/mol, 2 grams of hydrogen corresponds to 1 mole of H₂. The energy released when 2 grams of hydrogen is combusted is:
\[ \text{Energy released} = 2 \text{ g} \times -0.143 \text{ kJ/g} = -0.286 \text{ kJ} \]

### Step 5: Relate this to the enthalpy of formation of water
Since the combustion of 2 grams of hydrogen produces 2 moles of water, the enthalpy change for the formation of 1 mole of water is:
\[ \Delta H_f = \frac{-0.286 \text{ kJ}}{2} = -0.143 \text{ kJ/mol} \]
However, since we are looking for the enthalpy change for the formation of 1 mole of water, we need to consider the total energy released for the formation of 1 mole of water from its elements (H₂ and O₂).

### Step 6: Finalize the answer
Thus, the enthalpy of formation of water (ΔH_f) is:
\[ \Delta H_f = -286 \text{ kJ/mol} \]

### Conclusion
The ΔH_f of H₂O is -286 kJ/mol.

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