Question

Consider the following reaction:

Answer 1

Answer:

A) 0.83 atm

Explanation:

The given reaction is:

$CO2(g) + C(s)\rightleftharpoons 2CO(g)$

The equilibrium constant Kp is given as:

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

Initial [CO2] = 0.56 atm

Initial [CO] = 0.32 atm

Set-up ICE table

                         $CO2(g) + C(s)\rightleftharpoons 2CO(g)$

initial                          0.56                                              0.32

Change                      -x                                                   +2x

Eq                              0.56-x                                             0.32+2x

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

 

Answer 2

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= $\frac{ [0.32+x]^{2} }{ 0.56-x_{2} }$ 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.