Calculating K(sp) & Equilibrium Constants: A(g) --> 2B(g) + C(g)

[ChemRookie]

A(g) ----> 2B(g) + C(g)
<----

When 1.00mol of A is placed in a 4.00L container at temp t, the conventration of C at equilibrium is 0.050mol/L. What is the equilibrium constant for the reaction at temperature t?

Also have to Calculate the K(sp) for each of the salts whose solubility is lsited below...
a) CaSO4 = 3.3 * 10tothe-3mol/L
b) MgF2 = 2.7 * 10tothe-3mol/L

Thanks everyone.

 

[GCT]

[tex]K=\frac{^2[C]}{[A]} = \frac{[2x]^2[x]}{[.25M-x]} [/tex]

I'm not quite sure if the "2x" component is valid.

Once again, write out the actual Ksp equation for solubility and understand that each of the solubility values given are in term of the concentration when the solution is saturated with the compound, I believe it actually refers to the compound itself. If so, then you'll need to deduce the molarity of the each of the ionic components simply by using the formula. Multiply these two molarity components (using the proper exponents) to find Ksp.

 

[thepatientmental]

For the reaction A(g) --> 2B(g) + C(g), the equilibrium constant (K) can be calculated using the formula K = ^2[C]/[A]. In this case, we are given the initial concentration of A (1.00mol/4.00L = 0.250mol/L) and the equilibrium concentration of C (0.050mol/L). Plugging these values into the formula, we get K = (0.050)^2(0.050)/(0.250) = 0.005. Therefore, the equilibrium constant for this reaction at temperature t is 0.005.

For the salts whose solubility is listed, the K(sp) can be calculated using the formula K(sp) = [A]^m^n, where [A] and represent the equilibrium concentrations of the ions in the solution and m and n represent the coefficients of the ions in the balanced chemical equation.

a) For CaSO4, the balanced chemical equation is CaSO4(s) --> Ca2+(aq) + SO42-(aq). Therefore, the K(sp) for CaSO4 can be calculated as K(sp) = [Ca2+][SO42-]. Since the concentration of CaSO4 is given as 3.3 * 10^-3 mol/L, the equilibrium concentration of Ca2+ and SO42- will also be 3.3 * 10^-3 mol/L. Plugging these values into the formula, we get K(sp) = (3.3 * 10^-3)^2 = 1.089 * 10^-5.

b) For MgF2, the balanced chemical equation is MgF2(s) --> Mg2+(aq) + 2F-(aq). Therefore, the K(sp) for MgF2 can be calculated as K(sp) = [Mg2+][F-]^2. Since the concentration of MgF2 is given as 2.7 * 10^-3 mol/L, the equilibrium concentration of Mg2+ will also be 2.7 * 10^-3 mol/L. However, the concentration of F- will be twice that, since there are two moles of F- for every mole of MgF2. Thus, the equilibrium concentration of F- will be 2 * 2.7 * 10^-3 mol/L = 5.