100 mL of `PH_(3) on decomposition produced phosphorus and hydrogen. The change in volume is :-

Options

A. Increase in 50 ml
B. Decrease in 50 mL
C. Increase in 150 mL
D. Decrease in 200 ml

Answer

To solve the problem of determining the change in volume when 100 mL of PH₃ decomposes to produce phosphorus and hydrogen, follow these steps:

### Step 1: Write the balanced chemical equation for the decomposition of PH₃.
The balanced equation for the decomposition of phosphine (PH₃) is:
\[ 2 \text{PH}_3(g) \rightarrow 3 \text{H}_2(g) + 2 \text{P}(s) \]
Here, PH₃ decomposes to produce hydrogen gas (H₂) and solid phosphorus (P).

### Step 2: Analyze the volume relationship based on Avogadro's law.
According to Avogadro's law, the volume of a gas is directly proportional to the number of moles of gas at constant temperature and pressure. From the balanced equation, we can see that:
- 2 volumes of PH₃ produce 3 volumes of H₂.

### Step 3: Calculate the volume of H₂ produced from 100 mL of PH₃.
Using the stoichiometric relationship:
- If 2 volumes of PH₃ yield 3 volumes of H₂, then 1 volume of PH₃ yields \( \frac{3}{2} \) volumes of H₂.
- Therefore, 100 mL of PH₃ will yield:
\[ 100 \, \text{mL PH}_3 \times \frac{3 \, \text{mL H}_2}{2 \, \text{mL PH}_3} = 150 \, \text{mL H}_2 \]

### Step 4: Determine the change in volume.
The initial volume of PH₃ is 100 mL, and the volume of H₂ produced is 150 mL. The change in volume is calculated as:
\[ \text{Change in volume} = \text{Volume of H}_2 - \text{Volume of PH}_3 \]
\[ \text{Change in volume} = 150 \, \text{mL} - 100 \, \text{mL} = 50 \, \text{mL} \]

### Final Answer:
The change in volume after the decomposition of 100 mL of PH₃ is **50 mL**.

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