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 \( CO_2 \). We will also compare it with the heat liberated when 6 g of graphite is burnt.

### Step-by-Step Solution:

1. **Identify the Molar Mass of Carbon**:
The molar mass of carbon (C) is approximately 12 g/mol.

2. **Calculate Moles of Diamond**:
For 6 g of diamond:
\[ \text{Moles of Diamond} = \frac{\text{mass}}{\text{molar mass}} = \frac{6 \text{ g}}{12 \text{ g/mol}} = 0.5 \text{ mol} \]

3. **Understand the Reaction**:
The reaction for the combustion of diamond can be represented as:
\[ C_{\text{diamond}} + O_2 \rightarrow CO_2 \]
The enthalpy change for the conversion of diamond to graphite is given as \( \Delta H = -1.89 \text{ kJ} \). This means that when 1 mole of diamond is converted to graphite, 1.89 kJ of energy is released.

4. **Calculate Heat Released for 0.5 Moles of Diamond**:
Since the enthalpy change is for the conversion of 1 mole of diamond to graphite, we can calculate the heat released for 0.5 moles of diamond:
\[ \text{Heat released} = \Delta H \times \text{moles of diamond} = -1.89 \text{ kJ} \times 0.5 = -0.945 \text{ kJ} \]

5. **Conclusion**:
The heat liberated when 6 g of diamond is burnt is \( 0.945 \text{ kJ} \).