A pair of gases having same rate of diffusion at S.T.P

Options

A. CO , NO
B. `CO , CO_(2)`
C. `N_(2) O , NO_(2)`
D. `N_(2) O , CO_(2)`

Answer

To determine a pair of gases that have the same rate of diffusion at Standard Temperature and Pressure (S.T.P), we can utilize Graham's Law of Diffusion. This law states that at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Therefore, for two gases to have the same rate of diffusion, their molar masses must be equal.

### Step-by-Step Solution:

1. **Understand Graham's Law of Diffusion**:
- Graham's Law states that:
\[ \text{Rate of diffusion} \propto \frac{1}{\sqrt{\text{Molar mass}}} \]
- For two gases to diffuse at the same rate, their molar masses must be equal.

2. **List the Given Options**:
- We have four pairs of gases to evaluate:
- Option A: CO and NO
- Option B: CO and CO2
- Option C: N2O and NO2
- Option D: N2O and CO2

3. **Calculate the Molar Mass of Each Gas**:
- **Option A: CO and NO**
- Molar mass of CO = 12 (C) + 16 (O) = 28 g/mol
- Molar mass of NO = 14 (N) + 16 (O) = 30 g/mol
- **Conclusion**: Different molar masses.

- **Option B: CO and CO2**
- Molar mass of CO2 = 12 (C) + 2 × 16 (O) = 12 + 32 = 44 g/mol
- **Conclusion**: Different molar masses.

- **Option C: N2O and NO2**
- Molar mass of N2O = 2 × 14 (N) + 16 (O) = 28 + 16 = 44 g/mol
- Molar mass of NO2 = 14 (N) + 2 × 16 (O) = 14 + 32 = 46 g/mol
- **Conclusion**: Different molar masses.

- **Option D: N2O and CO2**
- Molar mass of N2O = 28 (from previous calculation)
- Molar mass of CO2 = 44 (from previous calculation)
- **Conclusion**: Different molar masses.

4. **Final Comparison**:
- After evaluating all options, we find that the only pair with the same molar mass is **N2O and CO2** which both equal 44 g/mol.

5. **Select the Correct Option**:
- Therefore, the correct answer is **Option D: N2O and CO2**.