Question

Calculate the partial pressure (in atm) of CH₃Br at equilibrium when 2.01 atm of CH₃OH and 2.01 atm of HBr react at 1000 K according to the following chemical equation:

CH3OH (g) + HBr (g) ?CH3Br (g) + H2O (g)

K_p = 2.78

Answer 1

                                CH₃OH (g) + HBr (g) CH₃Br (g) + H₂O (g)

initial pressure(atm)      2.01             2.01              0               0

change                        -a                  -a               +a             +a

Equb pressure         2.01-a           2.01-a            a                a

So Equilibrium constant ,

                     

                            

                                

                         

                              a = 5.27(2.01-a)

                                           a = 10.6 - 5.27a

                                    6.27 a = 10.6

                                           a = 1.69 atm

So the Equilibrium partial pressure of CH₃Br is = a = 1.69 atm

                                                                        a = 1.69x10⁰ atm   ( in scientific notation )