Question

10. The Kc value = 2.7 for the following water gas reaction at a given temperature:

                        CO(g) + H₂O(g)

Answer 1

Molecule                                Initial moles                             Final moles

CO                                        .13                                          .13 - x

H2O                                       .56                                          .56 - x

CO2                                       .62                                          .62 + x

H2                                          .43                                          .43 + x

a.

Kc = 2.7 = [(0.62+x)/5 *(0.43+x)/5] / [(0.13-x)/5 * (0.56-x)/5)}

Solving we get ,

x = dissociation constant = 1.73 (neglecting the negative value of x)

but from the value of x, final concentration of CO and H2O will be negative, hence the equilibrium will be towards the left.

b. Kp = Kc (RT)^z

where z is the difference in the number of moles of product and reactant = 0

hence Kp = Kc = 2.7

c.

Since quilibrium is towards left,

Molecule                                Initial moles                             Final moles

CO                                        .13                                          .13 + x

H2O                                       .56                                          .56 + x

CO2                                       .62                                          .62 - x

H2                                          .43                                          .43 - x

Keq = [(0.13+x)/5 *(0.56+x)/5] / [(0.62-x)/5 * (0.43-x)/5] = 2.7

hence x= .203

Equilibrium Concentration

CO = .13 + .2 = .15

H2O = .56 +.2 = .58

CO2 = .62 - .2 = .60

H2 = .43-.2 = .41                                                                                                                   < ANSWER>