150 mL of pure phosphine is decomposed to produce vapours of phosphorus and `H_2`. The change in volume during reaction is :

Options

A. 80mL
B. 75mL
C. 50mL
D. 60mL

Explanation

To solve the problem of the change in volume when 150 mL of pure phosphine (PH₃) decomposes to produce vapors of phosphorus (P₄) and hydrogen (H₂), we can follow these steps:

### Step 1: Write the balanced chemical equation for the decomposition of phosphine.
The decomposition reaction of phosphine can be represented as:
\[ 4 \text{PH}_3 \rightarrow \text{P}_4 + 6 \text{H}_2 \]

### Step 2: Determine the volume ratios from the balanced equation.
From the balanced equation, we see that:
- 4 moles of phosphine (PH₃) decompose to produce 1 mole of phosphorus (P₄) and 6 moles of hydrogen (H₂).
- Therefore, the volume ratios are:
- 4 mL of PH₃ produces 1 mL of P₄ and 6 mL of H₂.

### Step 3: Calculate the volume of hydrogen produced from the given volume of phosphine.
Given that we have 150 mL of PH₃, we can find the amount of hydrogen produced using the volume ratio:
\[ \text{Volume of H}_2 = \left(\frac{6 \text{ mL H}_2}{4 \text{ mL PH}_3}\right) \times 150 \text{ mL PH}_3 \]
\[ \text{Volume of H}_2 = \frac{6}{4} \times 150 = 225 \text{ mL} \]

### Step 4: Determine the change in volume during the reaction.
The change in volume can be calculated by subtracting the initial volume of phosphine from the total volume of gases produced (which includes the volume of hydrogen, since phosphorus is a solid and does not contribute to the gas volume):
\[ \text{Change in Volume} = \text{Volume of H}_2 - \text{Volume of PH}_3 \]
\[ \text{Change in Volume} = 225 \text{ mL} - 150 \text{ mL} = 75 \text{ mL} \]

### Conclusion:
The change in volume during the reaction is **75 mL**.

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