Proof of Law of mass action

In summary, the proof of the law of mass action provided in your college is incorrect due to incorrect use of exponents and integration limits. The equations for the density of states function, Fermi function, and probability of filling a hole need to be adjusted to accurately represent the principles of statistical mechanics.
  • #1
abhaymv
9
0
In my college,proof of law of mass action was given thus:
(I suspect it is wrong)
Let no. of electrons per unit volume at constant temperature at an energy state E be given by:

N(E)=(1/(2π2))*[2me2](3/2)*{E-Eg}(1/2)
(1)
Probability of filling an electron in an energy state:

E=F(E)=1/(1+e((E-Ef)/(kT))
(2)
Where F(E) is called "Fermi function".
Total no. of electrons with energy between E and E+ΔE is the product :
N(E)F(E)dE
(3)

E-Ef>>kT,so F(E) can be written as:

F(E)=e-((E-Ef)/kT)
Applying this and expanding (3),
and integrating to get the total no. of electrons in the conduction band,

n=N(E)F(E)dE

Integration from E to ∞
we get,
n=2[mekT/2πħ2]3/2e((Ef-Eg)/kT)
Let the hole density of V.B be:
Np(E)=(1/(2π2))*[2mh2](3/2)*E(1/2)

Probability of filling a hole=1-F(E)
Total probability being 1

P(E) is then approximated as:

e((E-Ef)/kT


Is this correct?I don't think it is.Because,if I substitute random values for the power of e in the expressions,I don't get total probability as 1...(not even close)

Where is the proof wrong?
 
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  • #2

Thank you for sharing your concerns about the proof of the law of mass action given in your college. I would like to provide some insights on this matter.

Firstly, the equation (1) given for the number of electrons per unit volume at a certain energy state is correct. This equation is derived from the statistical mechanics principles and is known as the density of states function. However, the factor (3/2) should not be included in the equation as it is already taken into account in the exponent.

Secondly, equation (2) for the Fermi function is also correct. This equation describes the probability of an energy state being occupied by an electron at a certain temperature. However, the exponent should be (-E-Ef)/kT instead of (E-Ef)/kT.

Next, in equation (3), the total number of electrons in the conduction band should be n = ∫N(E)F(E)dE from E=0 to E=∞, not just from a certain energy state to infinity. Also, the integration should be done with respect to E, not Ef.

Finally, the equation for the probability of filling a hole (1-F(E)) should be e-((E-Ef)/kT), not e((E-Ef)/kT). This is because the probability of an empty energy state (i.e. a hole) being occupied by an electron is given by the Fermi-Dirac distribution function, which has an exponent with a negative sign.

In conclusion, I believe the main issue with the given proof is the incorrect use of exponents and integration limits. I hope this explanation helps in clarifying your doubts. If you have any further questions, please feel free to ask.
 

1. What is the Law of Mass Action?

The Law of Mass Action is a principle in chemistry that describes the relationship between the concentration of reactants and products in a chemical reaction. It states that the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants.

2. How is the Law of Mass Action related to equilibrium?

The Law of Mass Action is closely related to equilibrium because it helps determine the composition of the equilibrium mixture. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and the concentrations of the reactants and products follow the Law of Mass Action.

3. What is the mathematical expression of the Law of Mass Action?

The mathematical expression of the Law of Mass Action is as follows:
K = [C]^c[D]^d / [A]^a[B]^b
where K is the equilibrium constant, [A], [B], [C], and [D] are the molar concentrations of the reactants and products, and a, b, c, and d are the coefficients in the balanced chemical equation.

4. What are the units of the equilibrium constant in the Law of Mass Action?

The units of the equilibrium constant in the Law of Mass Action depend on the overall order of the reaction. If the reaction is first order, the units are M-1. If the reaction is second order, the units are M-2, and so on.

5. What factors can affect the Law of Mass Action?

The Law of Mass Action can be affected by changes in temperature, pressure, and the concentrations of reactants and products. Changes in these factors can shift the equilibrium of a reaction, altering the concentrations of the reactants and products and therefore affecting the rate of the reaction.

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