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 problem regarding the effect of increased pressure on the equilibrium between diamond and graphite, we can follow these steps:

### Step 1: Understand the Equilibrium
The equilibrium in question is:
\[ C_{\text{diamond}} \rightleftharpoons C_{\text{graphite}} \]
This reaction represents the conversion of diamond to graphite and vice versa.

### Step 2: Identify the States of Matter
Both diamond and graphite are solids. In this equilibrium, we are dealing with two solid phases of carbon.

### Step 3: Apply Le Chatelier's Principle
Le Chatelier's Principle states that if an external change is applied to a system at equilibrium, the system will adjust to counteract that change. This principle is primarily applicable to changes in concentration, temperature, and pressure.

### Step 4: Consider the Effect of Pressure
When pressure is increased, the system will favor the side of the equilibrium that has fewer moles of gas. However, since both diamond and graphite are solids, they do not have a volume change associated with gas moles.

### Step 5: Conclusion
Since both reactants and products are solids and there is no change in the number of moles of gas, increasing the pressure will have no effect on the equilibrium position. Therefore, the equilibrium will remain unchanged.

### Final Answer
Increasing the pressure on the equilibrium \( C_{\text{diamond}} \rightleftharpoons C_{\text{graphite}} \) will have no effect on the position of the equilibrium.

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