Calculate the equilibrium concentrations of reactant and products when 0.329 moles of CO and 0.329 moles of Cl2 are introduced into a 1.00 L vessel at 600 K.
[CO] = ? M
[Cl2] = ? M
[COCl2] = ?M
I know how to create ICE tables but i get confused when it comes to doing the quadratic formula
The equilibrium constant, Kc, for the following reaction is 77.5 at 600 K. CO(g) + Cl2(g) COCl2(g)?
2018-03-07 10:13 am
回答 (2)
2018-03-07 4:46 pm
____________ CO(g) ______ + ______ Cl₂(g) ___ ⇌ ___ COCl₂(g) ___ Kc = 77.5
Initial: _____ 0.329 M ___________ 0.329 M _________ 0 M
Change: _____ -y M ______________ -y M __________ +y M
At eqm: __ (0.329 - y) M _______ (0.329 - y) M _______ y M
Kc = [COCl₂] / ([CO] [Cl₂])
77.5 = y / (0.329 - y)²
77.5 (0.329 - y)² = y
77.5 (y² - 0.658y + 0.108241) = y
77.5y² - 51.995y +8.3886775 = 0
For quadratic equation ax² + 2bx + c = 0, x = [-b ± √(b² - 4ac)] / 2a
Hence, y = [51.995 ± √(51.995² - 4×77.5×8.3886775)] / (2×77.5)
y = 0.270 or y = 0.401 (rejected, for 0.329 - 0.401 < 0)
At eqm :
[CO] = (0.329 - 0.270) M = 0.059 M
[Cl₂] = (0.329 - 0.270) M = 0.059 M
[COCl₂] = 0.270 M
Initial: _____ 0.329 M ___________ 0.329 M _________ 0 M
Change: _____ -y M ______________ -y M __________ +y M
At eqm: __ (0.329 - y) M _______ (0.329 - y) M _______ y M
Kc = [COCl₂] / ([CO] [Cl₂])
77.5 = y / (0.329 - y)²
77.5 (0.329 - y)² = y
77.5 (y² - 0.658y + 0.108241) = y
77.5y² - 51.995y +8.3886775 = 0
For quadratic equation ax² + 2bx + c = 0, x = [-b ± √(b² - 4ac)] / 2a
Hence, y = [51.995 ± √(51.995² - 4×77.5×8.3886775)] / (2×77.5)
y = 0.270 or y = 0.401 (rejected, for 0.329 - 0.401 < 0)
At eqm :
[CO] = (0.329 - 0.270) M = 0.059 M
[Cl₂] = (0.329 - 0.270) M = 0.059 M
[COCl₂] = 0.270 M
2018-03-07 10:44 am
...CO(g) + Cl2(g) = COCl2(g)
I...0.329 ,,,0.329 .......0
C,,,..-x ........-x ..........+x
E 0.329-x..0.329-x.....x
Kc = [COCl2(g)] / [CO(g)] * [Cl2(g)]
77.5 = x / .(0.329-x)^2
x = 77.5 ( .(0.329^2 -0.658 x + x^2)
x= 77.5 ( 0.108 - 0.658 x + x^2)
x= 8.389 - 50.995 x + 77.5x^2
77.5x^2 - 52x + 8.389 =0
x = 0.4010 and 0.2699
[CO] ={Cl2] = 0.329-0.2699 = 0.0590 M
[COCl2] = 0.270 M
I...0.329 ,,,0.329 .......0
C,,,..-x ........-x ..........+x
E 0.329-x..0.329-x.....x
Kc = [COCl2(g)] / [CO(g)] * [Cl2(g)]
77.5 = x / .(0.329-x)^2
x = 77.5 ( .(0.329^2 -0.658 x + x^2)
x= 77.5 ( 0.108 - 0.658 x + x^2)
x= 8.389 - 50.995 x + 77.5x^2
77.5x^2 - 52x + 8.389 =0
x = 0.4010 and 0.2699
[CO] ={Cl2] = 0.329-0.2699 = 0.0590 M
[COCl2] = 0.270 M
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