For the change,`C_("diamond") rarr C_("graphite"), Delta H = -1.89 kJ` , if 6 g of diamond and 6 g of graphite are separately burnt to yield `CO_2` the heat liberated in first case is:

Options

A. Less than in the second case by 1.89 kJ
B. Less than in the second case by 11.34 kJ
C. Less than in the second case by 14.34 kJ
D. More than in the second case by 0.945 kJ

Explanation

To solve the problem, we need to calculate the heat liberated when 6 g of diamond is burnt to yield carbon dioxide. We'll follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**: The combustion of diamond (C_diamond) can be represented as:
\[ C_{(diamond)} + O_2 \rightarrow CO_2 \]
The enthalpy change for the conversion of diamond to graphite is given as:
\[ \Delta H = -1.89 \, \text{kJ} \]

2. **Determine the Molar Mass of Carbon**: The molar mass of carbon (C) is 12 g/mol. Therefore, 12 g of carbon corresponds to 1 mole.

3. **Calculate Moles of Diamond**: Since we are burning 6 g of diamond, we calculate the number of moles of diamond:
\[ \text{Moles of diamond} = \frac{6 \, \text{g}}{12 \, \text{g/mol}} = 0.5 \, \text{mol} \]

4. **Relate the Enthalpy Change to Moles**: The enthalpy change of -1.89 kJ corresponds to 1 mole of diamond. For 0.5 moles of diamond, the heat liberated (ΔH1) can be calculated as:
\[ \Delta H_1 = \frac{-1.89 \, \text{kJ}}{1 \, \text{mol}} \times 0.5 \, \text{mol} = -0.945 \, \text{kJ} \]

5. **Conclusion**: The heat liberated when 6 g of diamond is burnt to yield carbon dioxide is:
\[ \Delta H_1 = -0.945 \, \text{kJ} \]

### Final Answer:
The heat liberated when 6 g of diamond is burnt to yield carbon dioxide is **0.945 kJ**.

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