Question

1-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.

2-Calculate the partial pressure (in atm) of H₂O at equilibrium when 1.38 atm of CH₃OH and 6.77 atm of HCl react at 2000 K according to the following chemical equation:

CH3OH (g) + HCl (g) ⇌ CH3Cl (g) + H2O (g)

Kp = 1.56

***Report your answer to three significant figures in scientific notation.

Answer 1

1. the given reaction is

-----------------------CdO(s) + H2(g) <------> Cd(s) + H2O(g), Kp = 4.12 x 10¹ (please conform the value)

eqm.prss(atm): 0, -----(3.21- P₀), -------- 0, ------ P₀

Kp = 4.12 x 10¹ = P(H2O,g) / P(H2,g) = P₀ / (3.21- P₀)

=> P₀ = 3.13x10¹ atm

Hence partial pressure due to H2O(g) at equilibrium = 3.13x10¹ atm (answer)

2: -------------------CH3OH (g) + HCl (g) <-----> CH3Cl (g) + H2O (g); Kp = 1.56

eqm.pres(atm):(1.38 - P₀), (6.77 - P₀) -------- P₀ ------------- P₀  

Kp = 1.56 = P₀ x P₀ / (1.38 - P₀)x(6.77 - P₀)

=> P₀ = 1.21x10¹ atm

Hence partial pressure due to H2O(g) at equilibrium = 1.21x10¹ atm (answer)