A reaction in equilibrium is respresnt by the following equation `2A_((s)) + 3B_((g))hArr3C_((g)) + D_((g)) + O_(2(g))` if the pressure on the system is reduced to half of its original value

Options

A. The amount of `C` "and ` D` decreases
B. The amounts of `C` "and" `D` increases
C. Decreasing the volume of the container
D. Increasing remain constant

Answer

To solve the problem, we will analyze the given equilibrium reaction and apply Le Chatelier's principle to determine the effect of reducing the pressure on the system.

### Step-by-Step Solution:

1. **Write the Balanced Equation:**
The reaction is given as:
\[ 2A_{(s)} + 3B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)} + O_{2(g)} \]
Here, \(A\) is a solid and \(B\), \(C\), \(D\), and \(O_2\) are gases.

2. **Identify the Gaseous Reactants and Products:**
- Gaseous reactants: \(3B_{(g)}\)
- Gaseous products: \(3C_{(g)} + D_{(g)} + O_{2(g)}\)
- Total moles of gaseous reactants = 3 (from \(B\))
- Total moles of gaseous products = 3 (from \(C\)) + 1 (from \(D\)) + 1 (from \(O_2\)) = 5

3. **Count the Moles of Gases:**
- Reactants: 3 moles of \(B\)
- Products: 5 moles (3 moles of \(C\) + 1 mole of \(D\) + 1 mole of \(O_2\))

4. **Apply Le Chatelier's Principle:**
According to Le Chatelier's principle, if the pressure of a system at equilibrium is decreased, the system will shift in the direction that produces more moles of gas to counteract the change.

5. **Determine the Direction of Shift:**
Since the number of gaseous moles increases from 3 (reactants) to 5 (products), the equilibrium will shift to the right (towards the products) when the pressure is reduced.

6. **Conclusion:**
As a result of the shift in equilibrium, the concentrations of \(C\), \(D\), and \(O_2\) will increase, while the concentration of \(B\) will decrease.

### Final Answer:
The amounts of \(C\) and \(D\) will **increase** when the pressure is reduced to half of its original value.

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