Question

Please show all work and all units

Using the following Equilibrium equation....

CO₂(g) + H_​2(g) <---> CO (g) + H2O (g)

InitIal pressures for all are 0.10 atm... Kc = 0.64 at room temperature

A) Calculate Kp value for rxn
​B) Calculate Rxn Qp
​C) To reach equilibrium, what direction must rxn proceed to reach equilibrium?
​D) Construct ICE Table for this equilibrium
​E) Solve for X and calculate the equilibrium pressures of each species.

Answer 1

(A) Given Equilibrium reaction is CO₂(g) + H_​2(g) <---> CO (g) + H2O (g)

We know that K_p = K_c x ( RT) ⁿ

Where

R = gas constant = 8.314 J/(mol-K)

T = Temperature in kelvin = 298 K

n = chnage in number of moles

        = Total number of moles of gaseous products - Total numner of moles of gaseous reactants

        = (1+1) - ( 1+1)

        = 0

Plug the values we get

K_p = 0.64 x () RT) ⁰

     = 0.64

(B)                         CO₂(g) + H_​2(g) <---> CO (g) + H2O (g)

initial pressure(atm)   0.10        0.10         0.10         0.10

Reaction quotient , Qc = (pCO x p H2O ) / ( p CO2 x pH2)

                                  = ( 0.10x0.10) / ( 0.10x 0.10)

                                  = 1

So Qc = 1

Since Qc > Kc so the reaction proceeds to the left

(C) In order to reach equilibrium the reaction proceeds towards right

(D)                            CO₂(g) + H_​2(g) <---> CO (g) + H2O (g)

Initial pressure (atm)    0.10       0.10        0.10         0.10

change                         -p           -p           +p          +p

Equb pressure         0.10-p      0.10-p      0.10+p    0.10+p

Equilibrium constant ,    Kp = 0.64 = (pCO x p H2O ) / ( p CO2 x pH2)

                                           0.64 = [(0.10+p)(0.10+p)]/[(0.10-p)(0.10-p)]

                      [(0.10+p)/(0.10-p)]² = 0.64

         [(0.10+p)/(0.10-p)] = 0.8

   0.10+p = 0.08 - 0.8p

     p = -0.011 atm

So Equilibrium pressure of CO₂(g) & H_​2(g) are = 0.10-p = 0.111 atm

Equilibrium pressure of CO (g) & H2O (g) are = 0.10+p = 0.089 atm