5

There are a few separate issues here to keep in mind: $K_c$ (the equilibrium constant in terms of concentrations) is defined as $$K_c = \prod {c_i}^{\nu_i} \tag{1}$$ Agreement between $K_c$ and the product of forward/back rate constants ($k_{\pu{fwd}}/k_{\pu{rev}}$ in the OP example) is expected only if the mechanism is correct, assuming some proportional ...


4

The concepts are extremely closely related. The difference is that the mass action ratio is a formally a ratio of concentrations, or, for gases, of partial pressures. The mass action ratio, and the term "mass action" in general, refers to situations where the chemical potential of the reactants and products is directly proportional to the concentration, or ...


3

This is a complicated question among chemical educationists- with tons of arguments over arguments. Just search do equilibrium constant have units? I would say the "units" of K are in a grey area (just like in any real world science.). You should read the nice section on Dimensionality of the Equilibrium Constant, and I will just give a background: Now, ...


3

T l, Dr: despite what you see from the aqueous electromotive series, potassium does not reduce calcium oxide in this setting. You may be assuming that potassium, which lies above calcium in the electromotive series for aqueous solutions, therefore lies above calcium in all other settings regardless of chemical environment or temperature. It doesn't. The ...


2

In a tight, solid vessel, you can of course have a liquid (or any condensed phase) completely fill it. There are basically three possibilities the inside is at (potentially very) high pressure. no problem, water is compressible, just not very much, and also the vessel has a finite $E$ modulus the inside is just at the vapour pressure of its content, but (...


2

The problem is that you are using wrong pressures. By definition for the reaction at equilibrium partial pressures can be expressed via initial partial pressure $P_1$ and conversion factor $α$ $$ \begin{array}{ccc} \ce{&A(g) &<=> &B(g) &+ &C(g)}\\ &(1 - α)P_1& & αP_1& & αP_1 \end{array} $$ equilibrium constant $...


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