Please show step by step Write the equilibrium-constant expression for the react
ID: 829591 • Letter: P
Question
Please show step by step
Write the equilibrium-constant expression for the reaction in terms of [A], [B], [C], and [D] as needed. For the reaction the equilibrium concentrations were found to be [AB2] = 0.360 M [B2] = 0.225 M [AB3] = 0.575 M What is the equilibrium constant for this reaction? Balance the equation and write the reaction-quotient expression, Qc. At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3. At this temperature, 0.300 mol of H2 and 0.300 mol of I2 were placed in a 1.00-L container to react. What concentration of HI is present at equilibrium? At a certain temperature, the equilibrium constant for the following chemical equation is 2.10. At this temperature, calculate the number of moles of N02(g) that must be added to 2.64 mol of SO3(g) in order to form 1.20 mol of S03(g) at equilibrium. Phosphorus pentachloride decomposes according to the chemical equation A 0.355 mol sample of PCl5(g) is injected into an empty 4.35 L reaction vessel held at 250 degree C. Calculate the concentrations of PCIs(g) and PC 13(g) at equilibrium.Explanation / Answer
1) Kc in terms of [A][B][C][D] will be [C]^2[D]/[A][B]^3 (Answer)
2) For the reaction equilibrium constant will be (0.575)^2/(0.225)(0.360)^2 = 11.338 (Answer)
3) Equation is U + 3F2 = UF6 (Answer)
and the reaction quotient expression Qc will be [UF6]/[U][F]^2 (Answer)
4) H2 + I2 <---> 2HI Kc for this reaction is 53.3
and 0.3 moles of H2 and I2 are placed in a 1 L container therefore
at equilibrium concentration of H2 I2 and HI will be 0.3-x , 0.3-x and 2x
Now Kc = [HI]^2/[H2][I2]
53.3 = 4x^2/(0.3-x)^2
On solvong we get x = 0.2355
therefore concentration of HI = 2x = 0.2355 x 2 = 0.471 (Answer)
5) So2 + No2 <---> So3 + No
Equilibrium constant is 2.1 = [S03][No]/[So2][No2]
If we add x moles of No2 then concentration will be
So2 + No2 <---> So3 + No
2.64-y x-y y y
Now y = 1.2 and K = 2.1
therefore
2.1 = [1.2][1.2]/[1.44][x-1.2]
therefore x = 1.67 moles (Answer)
6) PCl5 <----> PCl3 + Cl2 Kc = 1.8
0.355 moles of PCl5 are injected in 4.35L therefore Molarity = 0.355/4.35 = 0.0816
Now concentration of reactants will be
PCl5 <----> PCl3 + Cl2
0.0816-x x x
Now 1.8 = [x][x]/[0.0816-x]
On solving we get x = 0.0782
therefore
Concentration of PCl5 = 0.0816-x = 0.0034 (Answer)
and concentration of PCl3 = x = 0.0782 (Answer)