N₁(9)+3H₂(9)2NH,(9) At the temperature the engineer picks, the equilibrium constant K, for this reaction is 0.00014. The engineer charges ("fills") three reaction vessels with nitrogen and hydrogen, and lets the reaction begin. He then measures the composition of 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. reaction vessel compound pressure expected change in pressure N₂ 37.78 atm ↑ increase ○ ↓ decrease (no change) A Н2 46.78 atm ↑ increase ↓ decrease ○ (no change) NH3 23.31 atm ↑ increase O↓ decrease ○ (no change) Ny 33.67 atm ↑ increase ↓ decrease (no change) B H₂ 42.46 atm ↑ increase ↓ decrease NH3 17.18 atm ○ ↑ increase ↓ decrease (no change) ○ (no change) N₂ 33.58 atm ↑ increase ↓ decrease (no change) C H₂ 42.17 atm ↑ increase ○ ↓ decrease (no change) NH3 17.37 atm ↑ increase ○ ↓ decrease (no change)

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N2(g)+3H2(g) → 2 NH3(9)
At the temperature the engineer picks, the equilibrium constant K for this reaction is 0.00014.
Р
The engineer charges ("fills") three reaction vessels with nitrogen and hydrogen, and lets the reaction begin. He then measures the composition of 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.
reaction
vessel
compound
pressure
expected change in pressure
N₂
37.78 atm
↑ increase
↓ decrease
(no change)
A
H₂
46.78 atm
↑ increase
↓ decrease
(no change)
NH₁₂
23.31 atm
↑ increase
↓ decrease
(no change)
N₁₂
33.67 atm
↑ increase
↓ decrease
(no change)
B
H2
42.46 atm
↑ increase
↓ decrease
(no change)
NH₁₂
17.18 atm
↑ increase
↓ decrease
(no change)
N₂
33.58 atm
↑ increase
↓ decrease
(no change)
C
H2
42.17 atm
↑ increase
↓ decrease
(no change)
NH₁₂
17.37 atm
↑ increase
↓ decrease
(no change)
Transcribed Image Text:N2(g)+3H2(g) → 2 NH3(9) At the temperature the engineer picks, the equilibrium constant K for this reaction is 0.00014. Р The engineer charges ("fills") three reaction vessels with nitrogen and hydrogen, and lets the reaction begin. He then measures the composition of 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. reaction vessel compound pressure expected change in pressure N₂ 37.78 atm ↑ increase ↓ decrease (no change) A H₂ 46.78 atm ↑ increase ↓ decrease (no change) NH₁₂ 23.31 atm ↑ increase ↓ decrease (no change) N₁₂ 33.67 atm ↑ increase ↓ decrease (no change) B H2 42.46 atm ↑ increase ↓ decrease (no change) NH₁₂ 17.18 atm ↑ increase ↓ decrease (no change) N₂ 33.58 atm ↑ increase ↓ decrease (no change) C H2 42.17 atm ↑ increase ↓ decrease (no change) NH₁₂ 17.37 atm ↑ increase ↓ decrease (no change)
Suppose a 250. mL flask is filled with 1.8 mol of CO, 1.2 mol of CO2 and 0.50 mol of H2. This reaction becomes possible:
CO(g) + H2O(g) CO2(g) + H2(g)
=
Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the
equilibrium molarity of each compound after the reaction has come to equilibrium.
Use x to stand for the unknown change in the molarity of H2O. You can leave out the M symbol for molarity.
CO
H₂O
CO2
H₂
initial
1.8
-X
1.2
0.5
☑
change
1.2
x
+x
+x
equilibrium
0.2
X
1.2 + x
0.5 + x
Transcribed Image Text:Suppose a 250. mL flask is filled with 1.8 mol of CO, 1.2 mol of CO2 and 0.50 mol of H2. This reaction becomes possible: CO(g) + H2O(g) CO2(g) + H2(g) = Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of H2O. You can leave out the M symbol for molarity. CO H₂O CO2 H₂ initial 1.8 -X 1.2 0.5 ☑ change 1.2 x +x +x equilibrium 0.2 X 1.2 + x 0.5 + x
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