Calculate the change in pressure ( in atm) when 2 mole of `NO` and `16 g O _(2)` in `6.25` litre originally at `27^(@)C` react to produce the maximum quantity of `NO_(2)` possible according to the equation.
`( ` Take `R=(1)/(12)` ltr. Atm`//`mol K `)`
`2NO(g)+O_(2)(g) hArr 2NO_(2)(g)`

Options

A. 1
B. 4
C. 5
D. 2

Explanation

To solve the problem step by step, we will follow the given reaction and the information provided in the question.

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2 \text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO}_2(g) \]

### Step 2: Calculate the initial moles of reactants
- We are given 2 moles of NO.
- For O2, we need to calculate the number of moles from the given mass (16 g):
\[ \text{Moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 3: Determine the limiting reactant
From the balanced equation, 2 moles of NO react with 1 mole of O2. We have:
- 2 moles of NO
- 0.5 moles of O2

Since we have enough NO to react with the available O2, O2 is the limiting reactant.

### Step 4: Calculate the moles of products formed
When 0.5 moles of O2 react, they will produce:
\[ \text{Moles of } NO_2 = 2 \times \text{Moles of } O_2 = 2 \times 0.5 = 1 \text{ mole} \]

### Step 5: Calculate the remaining moles of reactants after the reaction
- Moles of NO remaining:
\[ \text{Initial moles of NO} - \text{Moles of NO used} = 2 - (2 \times 0.5) = 1 \text{ mole} \]
- Moles of O2 remaining:
\[ 0.5 - 0.5 = 0 \text{ moles} \]

### Step 6: Calculate the total moles after the reaction
The total moles after the reaction will be:
\[ \text{Moles of NO} + \text{Moles of NO}_2 = 1 + 1 = 2 \text{ moles} \]

### Step 7: Calculate the initial total moles
Initially, we had:
\[ \text{Initial moles} = \text{Moles of NO} + \text{Moles of O}_2 = 2 + 0.5 = 2.5 \text{ moles} \]

### Step 8: Calculate the change in number of moles
The change in number of moles (\(\Delta n\)) is:
\[ \Delta n = \text{Final moles} - \text{Initial moles} = 2 - 2.5 = -0.5 \text{ moles} \]

### Step 9: Use the ideal gas law to find the change in pressure
Using the ideal gas equation:
\[ PV = nRT \]
We can rearrange it to find the change in pressure (\(\Delta P\)):
\[ \Delta P = \frac{\Delta n \cdot R \cdot T}{V} \]
Where:
- \(R = \frac{1}{12} \, \text{L atm/(mol K)}\)
- \(T = 27^\circ C = 300 \, K\)
- \(V = 6.25 \, L\)

Substituting the values:
\[ \Delta P = \frac{-0.5 \cdot \frac{1}{12} \cdot 300}{6.25} \]

### Step 10: Calculate the change in pressure
Calculating this gives:
\[ \Delta P = \frac{-0.5 \cdot 25}{6.25} = \frac{-12.5}{6.25} = -2 \text{ atm} \]

### Final Answer
The change in pressure is:
\[ \Delta P = -2 \text{ atm} \]