Question

1-Calculate the partial pressure (in atm) of CH₃OH at equilibrium when 4.84 atm of CH₃I and 4.84 atm of H₂O react at 3000 K according to the following chemical equation:

CH3I (g) + H2O (g) ⇌ CH3OH (g) + HI (g)

Kp = 7.47×101

Report your answer to three significant figures in scientific notation.

2-Calculate the partial pressure (in atm) of H₂O at equilibrium when 59.6 g of CdO and 3.21 atm of H₂ react at 800 K according to the following chemical equation:

CdO(s) + H2(g) ⇌ Cd(s) + H2O(g)

Kp = 4.12×101

Report your answer to three significant figures in scientific notation.

Answer 1

Solution :-

1). Making the ICE table for the reaction

CH₃I (g)     +      H₂O (g) --- > CH₃OH (g) + HI (g)    Kp =7.47*10^1

4.84 atm            4.84                    0

-x                             -x                     +x

4.84-x                  4.84-x                 x

Kp = [CH3OH]/[CH3I][H2O]

7.47*10^1 = [x] /[4.84-x][4.84-x]

7.47*10^1 = x/[4.84-x]^2

Solving for x we get

X= 4.59

So the equilibrium pressure of the CH3OH = 4.59 atm

2).

Making ICE table

CdO(s)    +     H₂(g) ------ > Cd(s) + H₂O(g)             kp = 4.12*10^1

                        3.21 atm                         0

                        -x                                       +x

                        3.21-x                                x

Kp = [H2O]/[H2]

4.12*10^1 = [ x]/[3.21-x]

Solving for the x we get

X= 3.13 atm

So the partial pressure of the H2O2 = x = 3.13 atm