Question

Which pair of the gases diffuses with the same rate at same temperature and pressure? (a) \(\mathrm{CO}\) and \(\mathrm{NO}\) (b) \(\mathrm{NO}_{2}\) and \(\mathrm{CO}\) (c) \(\mathrm{NH}_{3}\) and \(\mathrm{PH}_{3}\) (d) \(\mathrm{NO}^{2}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}^{2}\)

Short answer

(a) and (c) are closest in pairs yet unrefined text, requiring option discussion for ideally CO and NO's definitions.

Step-by-step solution

Understanding Diffusion Rates

The rate of diffusion of gases is determined by Graham's law, which states that the rate of diffusion is inversely proportional to the square root of the molar mass of the gas. This means that two gases with equal molar masses will diffuse at the same rate.

Calculating Molar Masses

Calculate the molar masses for each of the gases: - CO (Carbon monoxide): Atomic mass of C = 12.01, O = 16.00, so Molar mass = 12.01 + 16.00 = 28.01 g/mol. - NO (Nitric oxide): Atomic mass of N = 14.01, O = 16.00, so Molar mass = 14.01 + 16.00 = 30.01 g/mol. - NO extsubscript{2} (Nitrogen dioxide): 14.01 + 2(16.00) = 46.01 g/mol. - CO extsubscript{2} (Carbon dioxide): 12.01 + 2(16.00) = 44.01 g/mol, but usually refers to ethane (wrongly interpreted initially as diatomic CO extsubscript{2}). - NH extsubscript{3} (Ammonia): 14.01 + 3(1.01) = 17.04 g/mol. - PH extsubscript{3} (Phosphine): 30.97 + 3(1.01) = 34.00 g/mol. - C extsubscript{2}H extsubscript{6} (Ethane): 2(12.01) + 6(1.01) = 30.08 g/mol.

Comparing Molar Masses

Look for pairs with equal molar masses: CO = 28.01 g/mol and C extsubscript{2}H extsubscript{6} = 30.08 g/mol, so not equal. NO = 30.01 g/mol which closely matches C extsubscript{2}H extsubscript{6} = 30.08 g/mol, but still not precisely equal. NH extsubscript{3} = 17.04 g/mol and PH extsubscript{3} = 34.00 g/mol, significantly different. NO extsubscript{2} = 46.01 g/mol and interpreted value requires careful check due to incorrect options. Based on closest initial mass, NO is best matched with, yet the problem should clarify.

Arriving at the Best Possible Option

Examine the options for the closest match: (a) CO and NO with close molar comparison but differing by 2 g/mol are not ideal, and (d) is incorrect without C extsubscript{2}H extsubscript{6} extsuperscript{2} correctly defined. None perfectly fit, typically such options require re-evaluation of text or typo. The closest correct option, based on problem setup, would depend on further refinement. Best identified as (a) CO and NO handling 'as listed' within closeness.

Key concepts

Molar Mass Calculation

Molar mass is a crucial concept when dealing with gases, especially when using Graham's law of diffusion. It represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To calculate the molar mass of a compound, you need to sum up the atomic masses of all the atoms present in that compound. For example, if you want to calculate the molar mass of carbon monoxide (CO), you start by identifying the atomic mass of carbon (12.01) and oxygen (16.00). Summing these values gives us a molar mass for CO of 28.01 g/mol. This process can be applied to any gas to determine its molar mass. Understanding molar mass helps us predict how gases will behave under different conditions and compare their diffusion rates.

Rates of Diffusion

The rate of diffusion for gases is a fascinating phenomenon described by Graham's law. This law posits that lighter gases diffuse more quickly than heavier ones because the rate of diffusion is inversely proportional to the square root of the molar mass. In simpler terms, if you have two gases with different molar masses, the gas with the lower molar mass will diffuse faster. For example, ammonia (NH extsubscript{3}) with a molar mass of 17.04 g/mol will diffuse more rapidly than phosphine (PH extsubscript{3}) which has a molar mass of 34.00 g/mol.Graham's law can be expressed mathematically as:\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]where \(r_1\) and \(r_2\) are the diffusion rates of gases 1 and 2, and \(M_1\) and \(M_2\) are their respective molar masses. This principle is essential in predicting how gases will spread in a given environment.

Gas Diffusion Experiment

Conducting a gas diffusion experiment is a practical way to observe Graham's law in action. Typically, this experiment involves two different gases, where you can measure how quickly each gas diffuses under controlled conditions of temperature and pressure. To conduct a simple gas diffusion experiment, follow these steps:

• Select two gases with different molar masses.
• Release them in a closed chamber from opposite ends.
• Use sensors or chemical methods to determine the point where both gases meet.
• Measure the diffusion path over time to calculate their rates of diffusion.

By comparing these rates, you can verify that lighter gases indeed diffuse faster than heavier ones. Such experiments are integral in teaching about molecular motion and the practical implications of Graham's law.