# Prove that Kc and Kp ONLY depend on Temperature Mathematically

I take A level Chemistry (Edexcel to be more precise), and my Chemistry teacher told me to 'Prove that Kc and Kp ONLY depend on Temperature Mathematically !'. After he kept on explaining how we do it ,I got really confused so my brain is Blank so could someone help me out. I searched ALL over the Internet to find something like what he asked but no results.

Thank you in Advance =]

KMST

Sat, 2011-10-08 11:35

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## I am not sure what your

I am not sure what your teacher meant. If he/she wants to use explanations assuming behavior of individual molecules in a reaction, it could get complicated. Any explanation would involve a model that, like models of the structure of an atom, could be debated or proven inadequate. I may be able to help you, or I may totally confuse you further. Science is confusing because we never know enough. The key to survival is not to get in deeper that you can handle, and to keep practicing your survival swimming.

Let's take the easy route first.

I take the chemical equilibrium law as such a well documented fact that it does not need proof, like the law of gravity. I believe that it has been proven that for a chemical reaction at a given temperature

aA +bB → cC + dD, the product of molar concentrations

K

_{c}= [A]^{a}[B]^{b}[C]^{-c}[D]^{-d }is a constant.I do not feel the need to prove that K

_{c}does not depend on anything other than temperature.K

_{P}is writen based on the partial pressures of the gaseous compounds involved in a reaction in gas phase. Assuming that the gases involved behave as ideal gases, we believe that they obey the ideal gas lawPV = nRT, so that the molar concentration of each gas (X) is

n

_{X}/V = P_{X}/(RT) = P_{X}(RT)^{-1}When we substitute that expression for the gases A, B, C, and D (X=A, X=B, and so on),

K

_{c}= [A]^{a}[B]^{b}[C]^{-c}[D]^{-d }converts intoK

_{c}= P_{A}^{a}P_{B}^{b}P_{C}^{-c}P_{D}^{-d}(RT)^{-a}(RT)^{-b}(RT)^{c}(RT)^{d}= P_{A}^{a}P_{B}^{b}P_{C}^{-c}P_{D}^{-d}(RT)^{(c+d-a-b)}= K_{P}(RT)^{-Δng}or K

_{P}= K_{c}(RT)^{Δng}where K

_{P}= P_{A}^{a}P_{B}^{b}P_{C}^{-c}P_{D}^{-d}and Δn_{g}= a+b-c-d = (a+b) - (c+d) is the change in the number of moles of gas in the reaction.So, in the simple view of the problem, K

_{c}is accepted as depending on nothing other than temperature as a law of physics/chemistry, and K_{P}is related to K_{c}in a way that does not introduce any other variable.The above may be enough for your teacher. Does it look like the explanation you were given?

If so, just to be contrary, I would like to point out that it is possible (if not likely) that a given reaction would have a K

_{c}independent of temperature within a certain temperature range. And if Δn_{g}= 0 for that reaction, K_{P}would be independent of temperature too.Now for more complicated explanations. Feel free to ignore what follows if your teacher is not asking for the complicated stuff.

The law of mass action was proposed based on the idea that reaction rates are proportional to products of the concentration of reactants. Since equilibrium is reached when the forward and backward reaction rates for a reaction are the same, the equation stating that both reaction rates are equal can be manipulated using algebra into the equilibrium constant expression.

We can define

Rate forward = k

_{f}[A]^{a}[B]^{b}, where k_{f}is a constant expected to depend on temperature onlyRate backward = k

_{B}[C]^{c}[D]^{d }, where k_{b}is a constant expected to depend on temperature onlyk

_{f}[A]^{a}[B]^{b}= k_{B}[C]^{c}[D]^{d}can be manipulated intok

_{b}/k_{f}= [A]^{a}[B]^{b}[C]^{-c}[D]^{-d }, so you say K_{c}=^{ }k_{b}/k_{f}is a constant that probably depends on temperature, but certainly depends on nothing else.If you have to explain why k

_{f}and k_{b}depend only on temperature, you could say that your model of a gas is tiny ball- like molecules bouncing around with speeds/kinetic energies dependent on temperature, and that the reactions are likely to occur when molecules of A and B (or C and D) collide. The chance of such a collision would only depend on the concentrations of the molecules and their speeds/kinetic energies. That model of gas behavior in a reaction would lead you to the reaction rates being proportional to the product of concentrations at any given temperature. The dependence on temperature, based on the fact that the molecules' speeds/kinetic energies depend on temperature, would be represented in k_{f}and k_{b}.thegoldengirl157

Sun, 2011-10-09 12:30

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## Thnak you for taking the time

Thnak you for taking the time to explain it to me in detail but it's not what my teacher asked for.

He said "I want you to show me your math skills " by showing that Kp and Kc ONLY depend on Temperature.

Anyways Thanks again =D