I do not understand this solution given by my professor, could someone please explain why Kc does not include concentrations of product or the other reactants? thank you.
For the following reaction at a 52°C, [C02] = 0.245 M. Calculate Kc and Kp. Mg(HC03)2 (s) 늑 MgO(s) H2O(l) 2 CO2(g) + + 2 C.
First, let us define the equilibrium constant for any species:
The equilibrium constant will relate product and reactants distribution. It is similar to a ratio
The equilibrium is given by
rReactants -> pProducts
Keq = [products]^p / [reactants]^r
For a specific case:
aA + bB = cC + dD
Keq = [C]^c * [D]^d / ([A]^a * [B]^b)
Where Keq is constant at a given temperature, i.e. it is dependant on Temperature only
If Keq > 1, this favours products, since this relates to a higher amount of C + D
If Keq < 1, this favours reactants, since this relates to a higher amount of A + B
If Keq = 1, this is in equilibrium, therefore, none is favoured, both are in similar ratios
Note that the concentrations MUST be in equilibrium. If these are not in equilibrium, then the reaction will take place until there is equilibrium achieved.
Typically; we use aqueous and gas phases. (Gases and Aqueous Concentrations can be related via PV = nRT, since M = n/V as well)
Solids and Liquids, i..e. (s) and (l) have activity of 1. Therefore, they must not be considered in the ratios.
Kc = [MgO] * [H2O] * CO2]^2 / [Mg(HCO3)2]
must be reduced to
Kc = [CO2]^2
and we know that
[CO2] = 0.245 M
Kc = (0.245)^2 = 0.060025 = 6*10^-2 approx
relate Kc and Kp
Kp = Kc*(RT)^dn
R= 0.082 ideal gas constant, T = 52 C= ,325K dn = mol of gas changes, in this case, 2 mol of gas produced, so dn = 2
Kp = 42.63