Densities of diamond and graphite are `3` and `2 g mL^(-1)`, respectively. The increase of pressure on the equilibrium `C_("diamond") hArr C_("graphite")`

Options

A. Favours backward reaction
B. Favours forwards reaction
C. Forms `3^(rd)` allotrope of carbon
D. increase the reaction rate

Answer

To solve the problem regarding the equilibrium between diamond and graphite under increased pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The equilibrium reaction is given as:
\[ C_{\text{diamond}} \rightleftharpoons C_{\text{graphite}} \]
This indicates that diamond and graphite are two allotropes of carbon in equilibrium with each other.

2. **Identify Densities**: We are given the densities:
- Density of diamond: \(3 \, \text{g/mL}\)
- Density of graphite: \(2 \, \text{g/mL}\)

3. **Analyze the Effect of Pressure**: According to Le Chatelier's principle, if we increase the pressure on a system at equilibrium, the system will shift in the direction that produces fewer moles of gas (or, in this case, the denser phase if both phases are solids).

4. **Comparing Densities**: Since the density of diamond (3 g/mL) is greater than that of graphite (2 g/mL), diamond is the denser form of carbon.

5. **Determine the Direction of Shift**: When pressure is increased, the equilibrium will shift towards the side with the higher density, which is diamond. Therefore, the reaction will favor the formation of diamond from graphite.

6. **Conclusion**: The increase in pressure will favor the backward reaction (the formation of diamond from graphite), thus shifting the equilibrium to the left.

### Final Answer:
The increase of pressure on the equilibrium \(C_{\text{diamond}} \rightleftharpoons C_{\text{graphite}}\) favors the backward direction (formation of diamond).