[electrogeek]

Hi everyone,

I’ve been asked to workout the energy of reaction at 0 kelvin for the reaction between HCl and DCl. The only thing I know is an absorption line of HCl of 2991 cm^-1.

From this I was able to work out the zero point energy for both HCl and DCl. I’ve presumed that they both have the same force constant, and the question states to use the approximation of a classical harmonic oscillator.

My zero point energy values for HCl and DCl are 17.9 KJ mol^-1 and 12.8 KJ mol^-1 respectively.

So would the answer simply be the difference between the zero point energies of HCl and DCl?

I’m not sure because I get an equilibrium constant of about 0.99 at 300 kelvin (this is when the energy of reaction is approximately the Gibbs free energy of the system) if I do treat the energy of reaction as the difference between the two, which seems quite high due to the amount of deuterium in the universe against the amount of hydrogen...

[mjc123]

What exactly is the reaction you're considering?

[electrogeek]

I was given the reaction HCl + D  DCl + H. However, I’ve checked my answer with enthalpy of formation values at 0 Kelvin and I got the same answer for both methods. Both gave a value of -5 KJ mol^-1 for the energy of reaction.

I have a different equilibrium constant now as well of about 7.6 when T = 300K using the assumption in the first message.

[Enthalpy]

Did you use the zero kelvin enthalpy of reaction to deduce the equilibrium constant at 300K?

The excited vibrational states of DCl and HCl are slightly populated at 300K (RT=2.5kJ/mol), which makes a small correction to the internal heats hence the enthalpy of reaction.

If you consider this correction unnecessary, it would be wise to explain why.

I almost forgot: what about the rotational energy at 300K? Is it close enough to 2/2RT for both DCl and HCl to neglect this contribution to the 300K enthalpy of reaction?

[electrogeek]

The question says to assume that they are equal for the calculation - it does go on to say that this isn't strictly true, and would be covered in later lectures....