At the temperature the engineer picks, the equilibrium constant K, for this reac
ID: 1026460 • Letter: A
Question
At the temperature the engineer picks, the equilibrium constant K, for this reaction is 24. The mixture inside each vessel from time to time. His first set of measurements are shown in the table below Predict the changes in the compositions the engineer should expect next time he measures the compositions. 02 so, H,o 2.49 am 3 94 atm 9 18 am | increase |O increase | decrease decrease decrease Q (no change) u(no change) (no change) | increase increase | 0 increase | 0 increase | increase decrease O decrease O decrease decrease (no change) d) (no change) (no change) G, (no change) o, | 225 atm 410 am 934 am 7.89 atm so. 5.59 atm increase O decrease (no change) 11.74 am increase O decrease (no change) oeExplanation / Answer
The balanced chemical equation for the reaction is
2 H2S (g) + 3 O2 (g) --------> 2 SO2 (g) + 2 H2O (g)
The equilibrium constant for the reaction is given as
We shall define the reaction quotient Q as below.
When Kp = Q, the reaction is at equilibrium. When Kp < Q, the reaction moves in the reverse direction to produce more of the reactants and attain the value of Kp. When Kp > Q, the reaction moves in the forward direction, favoring the products.
We determine the values of Q for the three cases below.
Reaction Vessel
Compound
Pressure (atm)
Q
Expected change in pressure
A
H2S
7.76
Q = (3.94)2(9.18)2/(7.76)2(2.49)3
= 1.4072
Kp > Q; to return to equilibrium, the value of the Q must increase. This is possible only when the pressure of the products increases. Therefore, we have
(i) pressure of H2S decreases.
(ii) pressure of O2 decreases.
(iii) pressure of SO2 increases.
(iv) pressure of H2O increases.
O2
2.49
SO2
3.94
H2O
9.18
B
H2S
7.60
Q = (4.10)2(9.34)2/(7.60)2(2.25)3
= 2.2289
Kp > Q; to return to equilibrium, the value of the Q must increase. This is possible only when the pressure of the products increases. Therefore, we have
(i) pressure of H2S decreases.
(ii) pressure of O2 decreases.
(iii) pressure of SO2 increases.
(iv) pressure of H2O increases.
O2
2.25
SO2
4.10
H2O
9.34
H2S
7.89
Q = (5.59)2(11.74)2/(7.89)2(1.43)3
= 23.65 24.0
Kp = Q; the reaction is at equilibrium and hence, the pressure of all the gaseous reactants and products will stay the same.
O2
1.43
SO2
5.59
H2O
11.74
Reaction Vessel
Compound
Pressure (atm)
Q
Expected change in pressure
A
H2S
7.76
Q = (3.94)2(9.18)2/(7.76)2(2.49)3
= 1.4072
Kp > Q; to return to equilibrium, the value of the Q must increase. This is possible only when the pressure of the products increases. Therefore, we have
(i) pressure of H2S decreases.
(ii) pressure of O2 decreases.
(iii) pressure of SO2 increases.
(iv) pressure of H2O increases.
O2
2.49
SO2
3.94
H2O
9.18
B
H2S
7.60
Q = (4.10)2(9.34)2/(7.60)2(2.25)3
= 2.2289
Kp > Q; to return to equilibrium, the value of the Q must increase. This is possible only when the pressure of the products increases. Therefore, we have
(i) pressure of H2S decreases.
(ii) pressure of O2 decreases.
(iii) pressure of SO2 increases.
(iv) pressure of H2O increases.
O2
2.25
SO2
4.10
H2O
9.34
H2S
7.89
Q = (5.59)2(11.74)2/(7.89)2(1.43)3
= 23.65 24.0
Kp = Q; the reaction is at equilibrium and hence, the pressure of all the gaseous reactants and products will stay the same.
O2
1.43
SO2
5.59
H2O
11.74
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