Well it’s the first answer couches
Most drugs are salt of either weak acids or weak bases they are absorbed in undissociated lipid soluble form. The dissociation if weak acids is suppressed by low (acid) PH. Thus theoretically one would expect a weak acid to be absorbed primarily in the stomach which has a low ph
According to that Kc is an equilibrium constant in terms of molar concentrations.
and Kc = [C]^c *[D]^d / [A]^a * [B]^b >>>> (1)
in the general reaction:
aA + bB ↔ cC + dD
and, from our balanced equation:
CH4 + H2O ⇔ Co + 3H2 >>> (2)
So, we need to calculate the concentrations (molarity) of the products and reactants:
the Molarity of CH4 = no. of moles/volume (L)
and no. of moles = weigh / Molecular weight = 42.3 / 16 = 2.643 moles
so the molarity of CH4 = 2.643 / 5 = 0.528 molar
the molarity of H20 = (49.2 / 18) / 5 = 0.546 molar
the molarity of CO = (8.32/28) / 5 = 0.059 molar
the molarity of H2 = (2.63 / 2) / 5 = 0.263 molar
By substitution in (1) according to (2);
∴ Kc = [0.059]*[0.263]^3 / ( [0.528]*[0.546]) = 3.7 * 10 ^-3 >>>> (3)
Kp = Kc (RT)^(Δn) >>> (4)
where R is the gas constant = 0.0821,
and Δn is the change in moles in gas= (3(H2) + 1 (CO) - (1 H2O + 1 CH4) = 2
by substition in (4):
∴ Kp = 3.7*10^-3 (0.0821* 1000)^2= 24.939
I believe 15 but i could be wrong
