A chemical system within a sealed 1 L reaction vessel is described by the following reversible reaction equation:
2H2S(g) <-->2H2(g) + S2(g)
If the equilibrium constant is 0.000 004 200 at 1103 K find:
a) the reaction quotient initially.
b) the order of concentration of all three components at equilibrium (greatest concentration to lowest) without using calculations.
c) what the size of the reaction quotient indicates regarding the extent of the forward reaction.
d) the equilibrium concentration of sulphur gas if 0.070 mol of hydrogen sulphide gas is initially placed in the vessel.
Note: for part d you may make the simplifying assumption that the amount of hydrogen sulphide gas reacted is extremely small in relation to the initial amount of hydrogen sulphide gas—this will allow you to avoid a quadratic expression in the denominator of the reaction quotient by disregarding one term).
The presentation of this question has serious flaws. Parts a, b and c can't be calculated since there are no values for anything in the question. If part d were done FIRST (not last) then all of the questions could be answered.
2H2S(g) <-->2H2(g) + S2(g)
a. I've never seen a -Q for a reaction with no values given for the reactants OR products. I have been told that if NO VALUES are given for reactants OR products that the Q cannot be determined. Or you can fill in the zero amounts like this; Qr = (0)^2(0)/(0)^2 =?
b. ..............2H2S(g) <-->2H2(g) + S2(g)
The equilibrium concentrations will depend upon the initial concentrations of at least ONE of the reagents in the equation. Since no values are given the equilibrium concentrations can't be calculated.
c. Since there is no Qr this question makes no sense.
d. If this question was made part a, then a, b and c could be calculated.
....................2H2S(g) <-->2H2(g) + S2(g)
I.....................0.070/L...........0...........0
C......................-2x...............2x...........x
E.................0.070-2x...........2x............x
K = 4.200E-6 = (H2)^2(S2)/(H2S)^2 = (2x)^2(x)/(0.070 - 2x)^2
If the 2x in 0.070-2x is ignored, as in the instructions, we have the following:
4.200E-6 = (2x)^2(x)/(0.070)^2 which is a cubic equation but can be done easily which I'll levave to you. Post your work if you get stuck.
Using part d first, the a, b and c have values. so can now be done.
a. Q = (H2)^2(S)/(H2S)^2 = (0)^2(0)/(0.70)^2 = 0
b. b can be done after you've solved part d.
c. It's unclear to me WHICH Q c refers.
This problem is so hard to interpret. I've done the best I can. If you post again and explain where I'm mixed up I will tackle it again.