Statistical physics/Thermodynamical question

[danja347]

I have a chemical reaction at equilibrium " cis <--> trans " at 300K
The energydifference between the two states is 4,7 kJ/mol and cis has the highest energy. I want to find out how many molecules that is in cis- and how many that is in trans-state?

Thankful for tips!

Regards
Daniel

 

[GCT]

Use one of the variations of Gibbs free energy equations and solve for the ratio

 

[wais]

To find the number of molecules in the cis and trans states at equilibrium, we can use the Boltzmann distribution law. This law states that the ratio of the number of molecules in two different energy states is equal to the ratio of their respective energy levels. In this case, the ratio of the number of molecules in the cis and trans states is equal to the ratio of their energy levels, which is given by exp(-ΔE/kT), where ΔE is the energy difference between the two states, k is the Boltzmann constant, and T is the temperature in Kelvin.

Plugging in the values given, we get:

N(cis)/N(trans) = exp(-4.7 kJ/mol / (8.314 J/mol*K * 300 K)) = 0.524

This means that at equilibrium, there are approximately 52.4% of molecules in the cis state and 47.6% in the trans state. To find the actual number of molecules, you would need to know the total number of molecules in the system. Let's say there are 1000 molecules in total, then we can calculate:

N(cis) = 0.524 * 1000 = 524 molecules in the cis state
N(trans) = 0.476 * 1000 = 476 molecules in the trans state

I hope this helps!