an ammonia molecule, NH3 travels at about 658 m/s at room temperature. How fast would a molecule of hydrogen sulfide, H2S travel under the same conditions?
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To determine the speed of a molecule of hydrogen sulfide (H2S) at room temperature, we need to consider the concept of root mean square (rms) speed. The formula for rms speed is given by:
vrms = sqrt(3kT/m)
Where:
- vrms is the root mean square speed of the molecule
- k is the Boltzmann constant (1.380649 x 10^-23 J/K)
- T is the temperature in Kelvin
- m is the mass of the molecule
First, let's calculate the temperature at room temperature, which is typically around 25 degrees Celsius or 298 Kelvin.
Next, we need to identify the molar mass of the ammonia (NH3) and hydrogen sulfide (H2S) molecules.
The molar mass of ammonia (NH3) is:
1 atom of Nitrogen (N) = 14.01 g/mol
3 atoms of Hydrogen (H) = 3.03 g/mol
Total = 17.04 g/mol
Convert the molar mass of ammonia to kilograms:
17.04 g/mol = 0.01704 kg/mol
The molar mass of hydrogen sulfide (H2S) is:
2 atoms of Hydrogen (H) = 2.03 g/mol
1 atom of Sulfur (S) = 32.07 g/mol
Total = 34.10 g/mol
Convert the molar mass of hydrogen sulfide to kilograms:
34.10 g/mol = 0.03410 kg/mol
Using the rms speed formula, we can calculate the speed of a molecule of hydrogen sulfide (H2S) at room temperature.
For ammonia (NH3):
vrms,NH3 = sqrt(3 * (1.380649 x 10^-23 J/K) * 298 K / 0.01704 kg/mol)
Similarly, for hydrogen sulfide (H2S):
vrms,H2S = sqrt(3 * (1.380649 x 10^-23 J/K) * 298 K / 0.03410 kg/mol)
Now, substitute the given values and calculate:
vrms,NH3 = sqrt(3 * (1.380649 x 10^-23 J/K) * 298 K / 0.01704 kg/mol)
vrms,NH3 ≈ 499 m/s (rounded to the nearest whole number)
vrms,H2S = sqrt(3 * (1.380649 x 10^-23 J/K) * 298 K / 0.03410 kg/mol)
vrms,H2S ≈ 412 m/s (rounded to the nearest whole number)
Therefore, a molecule of hydrogen sulfide (H2S) would travel at approximately 412 m/s under the same conditions as an ammonia (NH3) molecule traveling at 658 m/s.