Densities of diamond and graphite are `3.5` and `2.3 g mL^(-1)`, respectively. The increase of pressure on the equilibrium `C_("diamond") hArr C_("graphite")`

Options

A. Favours backward reaction
B. Fovours forward reaction
C. Have no effect
D. Increases the reaction rate

Explanation

To solve the question regarding the effect of pressure on the equilibrium between diamond and graphite, we can follow these steps:

### Step 1: Understand the Reaction
The equilibrium reaction is given as:
\[ C_{\text{diamond}} \rightleftharpoons C_{\text{graphite}} \]

### Step 2: Identify Densities
We are provided with the densities of diamond and graphite:
- Density of diamond = 3.5 g/mL
- Density of graphite = 2.3 g/mL

### Step 3: Analyze the Effect of Pressure
According to Le Chatelier's principle, if an external change (such as pressure) is applied to a system at equilibrium, the system will adjust to counteract that change.

### Step 4: Determine Which Form is Denser
In this case, diamond has a higher density (3.5 g/mL) compared to graphite (2.3 g/mL). This means that diamond is the denser form of carbon.

### Step 5: Apply Le Chatelier's Principle
When pressure is increased, the equilibrium will shift towards the side that has a smaller volume or, in this case, the denser product. Since diamond is denser than graphite, the equilibrium will shift to favor the formation of diamond.

### Conclusion
Thus, the increase of pressure on the equilibrium will favor the backward reaction (the formation of diamond from graphite).

### Final Answer
The correct option is: **Favors backward reaction.**

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