Calorific value of hydrogen gas is `- 143 kJ g^(-1)`. The standard enthalpy of formation of `H_(2)O ` will be

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Understand the Calorific Value**: The calorific value of hydrogen gas is given as -143 kJ/g. This means that when 1 gram of hydrogen gas is burned, it releases 143 kJ of energy.

2. **Determine the Reaction**: The standard enthalpy of formation of water can be represented by the reaction:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2O(l) \]
The enthalpy change for this reaction is what we want to find.

3. **Calculate the Enthalpy Change for 2 grams of H₂**: Since the molar mass of hydrogen (H₂) is approximately 2 g/mol, we will consider 2 grams of hydrogen for the calculation:
\[ \text{Energy released for 2 g of H}_2 = 2 \times (-143 \text{ kJ/g}) = -286 \text{ kJ} \]

4. **Relate the Heat of Combustion to the Enthalpy of Formation**: The heat of combustion for 2 grams of hydrogen is -286 kJ. Since the enthalpy of formation of water is defined as the energy change when forming 1 mole of water from its elements, we need to consider that the combustion of 2 grams of hydrogen produces 1 mole of water.

5. **Final Result**: Therefore, the standard enthalpy of formation of water (H₂O) is:
\[ \Delta H_f^\circ (\text{H}_2O) = -286 \text{ kJ/mol} \]

### Summary:
The standard enthalpy of formation of water (H₂O) is -286 kJ/mol.

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