To determine which gas will have the highest rate of effusion, we will use Graham's Law of Effusion. According to Graham's Law, the rate of effusion for a gas is inversely proportional to the square root of its molar mass. In simpler terms, lighter gases effuse faster than heavier gases.
### Pair 1: Oxygen ([tex]\( O_2 \)[/tex]) or Hydrogen ([tex]\( H_2 \)[/tex])
1. Find the molar mass of each gas:
- Oxygen ([tex]\( O_2 \)[/tex]): The atomic mass of oxygen (O) is approximately 16 amu. Since [tex]\( O_2 \)[/tex] has two oxygen atoms:
[tex]\[ \text{Molar mass of } O_2 = 2 \times 16 = 32 \text{ g/mol} \][/tex]
- Hydrogen ([tex]\( H_2 \)[/tex]): The atomic mass of hydrogen (H) is approximately 1 amu. Since [tex]\( H_2 \)[/tex] has two hydrogen atoms:
[tex]\[ \text{Molar mass of } H_2 = 2 \times 1 = 2 \text{ g/mol} \][/tex]
2. Compare the molar masses:
- The molar mass of [tex]\( O_2 \)[/tex] is 32 g/mol.
- The molar mass of [tex]\( H_2 \)[/tex] is 2 g/mol.
3. According to Graham's Law, the lighter gas [tex]\( (H_2) \)[/tex] will effuse faster.
Conclusion for Pair 1: Hydrogen ([tex]\( H_2 \)[/tex]) has the highest rate of effusion.
### Pair 2: Fluorine ([tex]\( F_2 \)[/tex]) or Chlorine ([tex]\( Cl_2 \)[/tex])
1. Find the molar mass of each gas:
- Fluorine ([tex]\( F_2 \)[/tex]): The atomic mass of fluorine (F) is approximately 19 amu. Since [tex]\( F_2 \)[/tex] has two fluorine atoms:
[tex]\[ \text{Molar mass of } F_2 = 2 \times 19 = 38 \text{ g/mol} \][/tex]
- Chlorine ([tex]\( Cl_2 \)[/tex]): The atomic mass of chlorine (Cl) is approximately 35.5 amu. Since [tex]\( Cl_2 \)[/tex] has two chlorine atoms:
[tex]\[ \text{Molar mass of } Cl_2 = 2 \times 35.5 = 71 \text{ g/mol} \][/tex]
2. Compare the molar masses:
- The molar mass of [tex]\( F_2 \)[/tex] is 38 g/mol.
- The molar mass of [tex]\( Cl_2 \)[/tex] is 71 g/mol.
3. According to Graham's Law, the lighter gas [tex]\( (F_2) \)[/tex] will effuse faster.
Conclusion for Pair 2: Fluorine ([tex]\( F_2 \)[/tex]) has the highest rate of effusion.
