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Consider the following reaction:

CO2(g) + C(graphite) ⇌ 2 CO(g)

A reaction mixture initially contains 0.56 atm CO2 and 0.32 atm CO. Determine the equilibrium pressure of CO if Kp for the reaction at this temperature is 2.25.
A) 0.83 atm
B) 0.31 atm
C) 0.26 atm
D) 0.58 atm
E) 0.42 atm

Respuesta :

The Kp in this problem is expressed as Kp= [CO]^2/[CO2] and is equal to 2.25. Substituting the initial concentrations we get Kp= [tex] \frac{ [0.32+x]^{2} }{ 0.56-x_{2} } [/tex] where x is the amount that is produced. x s calculated is 0.255 atm. Thus, the final partial pressure of CO is 0.83 atm. Answer is A.

Answer:

A) 0.83 atm

Explanation:

The given reaction is:

[tex]CO2(g) + C(s)\rightleftharpoons 2CO(g)[/tex]

The equilibrium constant Kp is given as:

[tex]Kp = \frac{[CO]^{2} }{[CO2]}[/tex]

Initial [CO2] = 0.56 atm

Initial [CO] = 0.32 atm

Set-up ICE table

                         [tex]CO2(g) + C(s)\rightleftharpoons 2CO(g)[/tex]

initial                          0.56                                              0.32

Change                      -x                                                   +2x

Eq                              0.56-x                                             0.32+2x

[tex]2.25 = \frac{(0.32+2x)^{2} }{(0.56-x)} \\\\x = 0.254 atm\\\\Equilibrium\ pressure\ CO = 0.32 + 2x = 0.32 + 2(0.254) =0.81 atm[/tex]

 

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