1-Calculate the partial pressure (in atm) of H 2 O at equilibrium when 59.6 g of
ID: 986092 • Letter: 1
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:
***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:
***Report your answer to three significant figures in scientific notation.
CdO(s) + H2(g) Cd(s) + H2O(g) Kp = 4.12×101Explanation / Answer
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)
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