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What would be the relative population of the J = 1 and J = 0 rotational states of HCl at 25°C_______?
(Given: BHCl = 10.591 cm–1) [Rounded up to first decimal place]
(Given: BHCl = 10.591 cm–1) [Rounded up to first decimal place]
Correct answer is between '2.6,2.8'. Can you explain this answer?
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What would be the relative population of the J = 1 and J = 0 rotationa...
°C if the total number of molecules in the sample is 1.00 × 10^16?
To determine the relative population of the J = 1 and J = 0 rotational states of HCl at 25°C, we can use the Boltzmann distribution equation:
N(J)/N = (2J + 1) exp(-E(J)/kT)
where N(J) is the number of molecules in the J rotational state, N is the total number of molecules in the sample, E(J) is the energy of the J rotational state, k is the Boltzmann constant, and T is the temperature in Kelvin.
For HCl, the energy of the J = 1 rotational state is E(1) = (1/2)hB, where h is the Planck constant and B is the rotational constant. The energy of the J = 0 rotational state is E(0) = 0.
At 25°C, T = 298 K. The rotational constant of HCl is B = 10.593 cm^-1.
Plugging in the values, we get:
N(1)/N = 3 exp(-E(1)/kT) = 3 exp(-hB/2kT) = 3 exp(-2.93)
N(0)/N = 1 exp(-E(0)/kT) = 1 exp(0) = 1
Thus, the relative population of the J = 1 and J = 0 rotational states of HCl at 25°C would be:
N(1)/N(0) = 3 exp(-2.93) = 0.003
This means that the J = 1 state is only about 0.3% as populated as the J = 0 state at 25°C.
To determine the relative population of the J = 1 and J = 0 rotational states of HCl at 25°C, we can use the Boltzmann distribution equation:
N(J)/N = (2J + 1) exp(-E(J)/kT)
where N(J) is the number of molecules in the J rotational state, N is the total number of molecules in the sample, E(J) is the energy of the J rotational state, k is the Boltzmann constant, and T is the temperature in Kelvin.
For HCl, the energy of the J = 1 rotational state is E(1) = (1/2)hB, where h is the Planck constant and B is the rotational constant. The energy of the J = 0 rotational state is E(0) = 0.
At 25°C, T = 298 K. The rotational constant of HCl is B = 10.593 cm^-1.
Plugging in the values, we get:
N(1)/N = 3 exp(-E(1)/kT) = 3 exp(-hB/2kT) = 3 exp(-2.93)
N(0)/N = 1 exp(-E(0)/kT) = 1 exp(0) = 1
Thus, the relative population of the J = 1 and J = 0 rotational states of HCl at 25°C would be:
N(1)/N(0) = 3 exp(-2.93) = 0.003
This means that the J = 1 state is only about 0.3% as populated as the J = 0 state at 25°C.
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What would be the relative population of the J = 1 and J = 0 rotational states of HCl at 25°C_______? (Given: BHCl = 10.591 cm–1) [Rounded up to first decimal place]Correct answer is between '2.6,2.8'. Can you explain this answer?
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