I will reward the most informative answer. 1. When 0.152 mol of solid PH 3 BCl 3
ID: 791429 • Letter: I
Question
I will reward the most informative answer.
1. When 0.152 mol of solid PH3BCl3 is introduced into a 3.0 L container at a certain temperature, 8.44
I will reward the most informative answer. When 0.152 mol of solid PH3BCl3 is introduced into a 3.0 L container at a certain temperature, 8.44 ´10-3 mol of PH3 is present at equilibrium: PH3BCl3(s) PH3(g) + BCl3(g) Construct a reaction table for the process, and use it to calculate Kc at this temperature. 10.0 mL of a 0.100 mol L-1 solution of a metal ion M2+ is mixed with 10.0 mL of a 0.100 mol l-1 solution of a substance L. The following equilibrium is established: M2+(aq) + 2L(aq) ML22+(aq) At equilibrium the concentration of L is found to be 0.0100 mol L-1. What is the equilibrium concentration of ML22+, in mol L-1? Consider the reactions of cadmium with the thiosulfate anion. Cd2+(aq) + S2O32-(aq) Cd(S2O3)(aq) K1 = 8.3 ´ 103 Cd(S2O3)(aq) + S2O32-(aq) Cd(S2O3)22-(aq) K2 = 2.5 ´ 102 What is the value for the equilibrium constant for the following reaction? Cd2+(aq) + 2S2O32-(aq) Cd(S2O3)22-(aq) The reaction system POCl3(g) POCl(g) + Cl2(g) is at equilibrium. Which of the following statements describes the behavior of the system if POCl is added to the container?Explanation / Answer
We have : PH3BCl3(s) <==> PH3(g) + BCl3(g)
Initial : 0.152 mol 0 0
Change: -x +x +x
Final: 0.152-x x x
We know that final amount of PH3 = 8.44*10^-3 mol
So, x = 8.44*10^-3 mol
Now, Kc = [PH3] [BCl3]
Now, [PH3] = {BCl3] = 8.44'10^-3 / 3 [since Volume of the container is 3L]
Kc = [2.81*10^-3] ^2 = 7.91 * 10^-6 [This value shows that the reaction hardly goes in the forwrds direction]
Also, we have not taken the reactant to calculate the value of Kc as the reactant is in solid state.
2) Consider : M2+(aq) + 2L(aq) <==> ML22+(aq)
Initial moles: 0.001 0.001 0
change moles: -x -2x +x
Final moles: (0.001-x) (0.001-2x) x
Now finally we have 0.01 molL-1 in 20 ml i.e 0.0002 moles
so, 0.001-2x = 0.0002
x = 0.0004 moles in 20ml = 0.02mol L-1
So concentration of product = 0.02 mol/L
3) Cd2+(aq) +S2O32