For the reaction 2A(g) + B(aq) + 3C(l)<------> D(s) + 3E(g), the concentrations at equilibrium are found to be:
A: 2.3x10^3 Pa
B: 1.8x10^-2 M
C: 15.8M
D: 12.5M
E: 5.6x10^4 Torr
Find the numerical value of the equilibrium constant that would appear in a conventional table of equilibrium constants.
The problem is misleading. It says the numbers "below" are concentrations but they aren't. A is in pascals; although pressure and M can be mixed, pascals must be converted to atmospheres. 101325 Pa = 1 atm. Likewise, E is in torr and must be converted to atm. 760 torr = 1 atm.
To find the numerical value of the equilibrium constant, we need to set up the equilibrium expression for the given reaction and determine the stoichiometric coefficients.
The equilibrium constant expression is written as follows:
Kc = [D]^a [E]^b / [A]^c [B]^d [C]^e
Where [D], [E], [A], [B], [C] represent the molar concentrations of each species at equilibrium, and a, b, c, d, e are the stoichiometric coefficients of each species in the balanced equation.
From the balanced equation, we can determine the stoichiometric coefficients:
2A(g) + B(aq) + 3C(l) <------> D(s) + 3E(g)
a = 1 (for D)
b = 3 (for E)
c = 2 (for A)
d = 1 (for B)
e = 3 (for C)
Now we can substitute the given concentrations into the equilibrium expression and solve for Kc.
Kc = [D]^1 [E]^3 / [A]^2 [B]^1 [C]^3
Since the concentration of D, E, A, B, and C are given:
[D] = 12.5 M
[E] = 5.6x10^4 Torr = (5.6x10^4 Torr) / (760 Torr/atm) ≈ 73.68 atm
[A] = 2.3x10^3 Pa = (2.3x10^3 Pa) / (1 atm/101325 Pa) ≈ 2.27x10^-2 atm
[B] = 1.8x10^-2 M
[C] = 15.8 M
Substituting these values into the equilibrium expression:
Kc = (12.5)(73.68^3) / (2.27x10^-2)^2 (1.8x10^-2)(15.8^3)
Simplifying the expression will yield the numerical value of Kc.
Solids and liquids don't count for Keq; therefore,
Keq = (E)^3/(A)^2(B)
Substitute those values from the problem and solve for Keq.