Given reaction is
2CO2(g)
2CO(g) + O2(g)
Lets see the initial concentration of CO2
[CO2 ] = 2.0 mol / 5.0 L = 0.4 M
now lets see the ICE table for given reaction
...................................2CO2(g)..............2CO(g)......+.....O2(g)
Initial.......................0.4 M........................0....................0
Change ..................-2x............................+2x..................+x
At Equilibrium........0.4-2x.........................2x....................x
Now equilibrium constant for reaction is
Kc = [CO]2[O2] /[CO2]2
2.00 10-4
= (2x)2
(x) /
(0.4-2x)2 = 4x3
/ (0.4-2x)2
as x <<< 0.4 we can neglect 0.4-2x = 0.4 then
2.00 10-4
= 4x3 / (0.4)2 = 4x3 /
0.16
x3 = 0.16 2.00
10-4 / 4 = 0.08
10-4
Lets take the cube root on both the sides we get
x = 0.02 M
then at equilibrium
[CO] = 2x = 2 0.02 = 0.04
M
[O2]= x = 0.02 M
[CO2] = 0.4 -2x = 0.4 - 2 0.02 = 0.36
M
Calculate the equilibrium concentrations of all species. (Use Kc equilibrum equation). Please show work, thank you!...
methor eter, r 1 13 Vocz ANP 151 Exam 2 Version W (2002(g) 209(9) + O2(g) Ks = 2.00x 10 If 2.0 mol CO2 is initially placed into a 5.0 L vessel, calculate the equilibrium concentrations of all species.
show all work and equations used pls
2.50 mol CO2 was initially placed into a 1.00-L container and allowed to come to equilibrium. 2CO2(8) = 2CO(g) + O2(8) At equilibrium, 0.95 mol CO2 remains. What is the value of Kc?
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s Principle [References) At a particular temperature, K = 7.0 x 10-6 for the reaction 2 CO2(g) = 2 CO(g) + O2(g) If 4.0 moles of CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. (CO2) = (CO) = [02] = Submit Answer Try Another Version 9 item attempts remaining
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Carbon dioxide reacts to form carbon monoxide and oxygen by the following equation: 2CO2 -> 2CO + O2; K= 2.00 x 10-6 mol/L If 3.00 mmol CO2 is initially placed into a 5.00 L chamber, calculate the equilibrium concentration of all three species. Use the equilibrium ICE table and the 5% rule.
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you!
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