The system
CO2(g) + H2(g) ⇌ H2O(g) + CO(g)
is at equilibrium at some temperature. At equilibrium, a 4.00L vessel contains 1.00 mole CO2, 1.00 mole H2, 2.40 moles H2O, and 2.40 moles CO. How many moles of CO2 must be added to this system to bring the equilibrium CO concentration to 0.669 mol/L?
0.993 moles
0.498 moles
0.069 moles
0.429 moles
Moles question & equilibrium?
2016-02-27 5:33 pm
回答 (2)
2016-02-27 6:16 pm
CO₂(g) + H₂(g) ⇌ H₂O(g) + CO(g) ... Kc = [H₂O][CO]/([CO₂][H₂])
In the original equilibrium state :
[CO₂] = [H₂] = 1.00/4.00 = 0.250 mol/L
[H₂O] = [CO] = 2.40/4.00 = 0.600 mol/L
Suppose 4y mol of CO₂ is added. When the new equilibrium state is established :
[CO] = 0.669 mol/L
Increase in concentration of CO = 0.669 - 0.600 = 0.069 mol/L
[H₂O] = 0.600 + 0.069 = 0.669 mol/L
[H₂] = 0.250 - 0.069 = 0.181 mol/L
[CO] = 0.250 + y - 0.069 = y + 0.181 mol/L
At fixed temperature, Kc is constant.
0.600²/0.250² = 0.669²/{(y + 0.181) × 0.181}
y + 0.181 = (0.669²/0.181) × (0.250²/0.600²)
y = (0.669²/0.181) × (0.250²/0.600²) - 0.181
4y = 4 × {(0.669²/0.181) × (0.250²/0.600²) - 0.181}
4y = 0.993
No. of moles of CO₂ must be added = 0.993 mol
In the original equilibrium state :
[CO₂] = [H₂] = 1.00/4.00 = 0.250 mol/L
[H₂O] = [CO] = 2.40/4.00 = 0.600 mol/L
Suppose 4y mol of CO₂ is added. When the new equilibrium state is established :
[CO] = 0.669 mol/L
Increase in concentration of CO = 0.669 - 0.600 = 0.069 mol/L
[H₂O] = 0.600 + 0.069 = 0.669 mol/L
[H₂] = 0.250 - 0.069 = 0.181 mol/L
[CO] = 0.250 + y - 0.069 = y + 0.181 mol/L
At fixed temperature, Kc is constant.
0.600²/0.250² = 0.669²/{(y + 0.181) × 0.181}
y + 0.181 = (0.669²/0.181) × (0.250²/0.600²)
y = (0.669²/0.181) × (0.250²/0.600²) - 0.181
4y = 4 × {(0.669²/0.181) × (0.250²/0.600²) - 0.181}
4y = 0.993
No. of moles of CO₂ must be added = 0.993 mol
2016-02-27 6:15 pm
good question, but tough. First calculate Kc from the equilibrium given: [CO2] = [H2] = 0.250 M and [H2O] = [CO] = 0.600 M
now add 4y mol CO2 so the new initial conditions have [CO2] = 0.250 + y, [H2] = 0.250 M, and [H2O] = [CO] = 0.600 M
Now use an ICE for this new initial conditions. The eqm concentrations are
[CO2] = 0.250 + y -x
[H2] = 0.250 - x
[H2O] = 0.600 + x
[CO] = 0.600 + x
where 0.600 + x = 0.669 M, so calculate x, you know Kc already, so the only unknown is y. Solve for y and the #mol needed to add is 4y
now add 4y mol CO2 so the new initial conditions have [CO2] = 0.250 + y, [H2] = 0.250 M, and [H2O] = [CO] = 0.600 M
Now use an ICE for this new initial conditions. The eqm concentrations are
[CO2] = 0.250 + y -x
[H2] = 0.250 - x
[H2O] = 0.600 + x
[CO] = 0.600 + x
where 0.600 + x = 0.669 M, so calculate x, you know Kc already, so the only unknown is y. Solve for y and the #mol needed to add is 4y
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