[emissivity]

maybe you can see the picture.

I just want to derive the secular equation.

So I've got the energy equation and

I used the fact that its partial derivatives are 0.

but I don't know why there is ''S'' and ''E''

and why its denominator is not square.



please help me..

otherwise, I just have to memorize the secular equation..




ahhhh
these are the problems from '''''Atkins' physical chemistry 9th ed, peter atkins, oxford, p390~391''''

[Schrödinger]

Instead of applying the u/v rule, try taking the denominator to the left hand side and use the u.v rule. This is easier

[juanrga]

Setting
$$E = \frac{U}{V}$$
the partial derivative is
$$\frac{\partial E}{ \partial c_A} = \frac{1}{V} \frac{\partial U}{ \partial c_A} - \frac{U}{V^2} \frac{\partial V}{ \partial c_A}$$
Differentiating the numerator ##U## we obtain ##2 c_A \alpha_A + 2 c_B \beta## and both terms are found in the answer but are not found in your attempt. I do not know what you really did.

[emissivity]

your comment is really helpful.

I made it!



dear Juanrga,

thanks for your comment,
but I think Schrodinger's way is easier..

thank you!

[juanrga]

No problem you would select the answer that better suit your needs and tastes. Let me show why I think that my method is easy

The above expression for the derivative of a fraction ##E=U/V## can be written in the equivalent form
$$\frac{\partial E}{ \partial c_A} = \frac{1}{V} \left( \frac{\partial U}{ \partial c_A} - E \frac{\partial V}{ \partial c_A} \right)$$
Differentiating the numerator
$$\frac{\partial U}{ \partial c_A} = 2 c_A \alpha_A + 2 c_B \beta$$
Differentiating the denominator
$$\frac{\partial V}{ \partial c_A} = 2 c_A + 2 c_B S$$
Substituting
$$\frac{\partial E}{ \partial c_A} = \frac{2 (c_A \alpha_A + c_B \beta - c_A E - c_B SE)}{V}$$
which is exactly the answer in the textbook.

In my opinion this method is not more complex than moving the denominator in the original fraction to the left hand side, next use the u·v rule, later move the E part to the right hand side, and finally divide both left and right hand sides by the original denominator; but others can disagree of course!