Question

For \( \text{H}_2 \) molecule, the fundamental vibrational frequency \( \nu_e \) can be taken as 4400 cm\(^{-1}\). The zero-point energy of the molecule is ................ kJ/mol (rounded up to two decimal places). 
\[\left[ h = 6.6 \times 10^{-34}\,\mathrm{J\,s}, \; c = 3 \times 10^{8}\,\mathrm{m\,s^{-1}}, \; N_A = 6 \times 10^{23}\,\mathrm{mol^{-1}} \right]\]

 

Hint:

The zero-point energy of a molecule can be calculated using \( E_0 = \frac{1}{2} h \nu_e \), where \( \nu_e \) is the fundamental vibrational frequency.

Answer 1

Correct Answer: 25.8 - 26.4

Step 1: Formula for zero-point energy. 
The zero-point energy (\( E_0 \)) for the molecule is given by: \[ E_0 = \frac{1}{2} h \nu_e \] where \( \nu_e = 4400 \, \text{cm}^{-1} \). First, we convert this to SI units (Hz) using the speed of light: \[ \nu_e = 4400 \, \text{cm}^{-1} \times c = 4400 \times 3 \times 10^{10} = 1.32 \times 10^{14} \, \text{Hz} \]

Step 2: Calculating the zero-point energy. 
Now, substitute the values into the formula: \[ E_0 = \frac{1}{2} \times 6.626 \times 10^{-34} \times 1.32 \times 10^{14} = 4.38 \times 10^{-20} \, \text{J} \] To convert this to kJ/mol, we multiply by \( N_A \) and divide by 1000: \[ E_0 = \frac{4.38 \times 10^{-20} \times 6 \times 10^{23}}{1000} = 26.28 \, \text{kJ/mol} \]

Step 3: Conclusion. 
The zero-point energy of the \( \text{H}_2 \) molecule is 26.28 kJ/mol.