{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Assignemnt 7 Key - Question:1 The first equilibrium...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Question:1. The first equilibrium expression (K C ) as represented by the first chemcial equation is 5.81 × 10 -7 at 700 K. Calculate the value of the equilibrium expression for the second equation at the same temperature. Solution: The equilibrium expression for is, The equilibrium expression for the second equation, , is Notice the relationship between the two equilibrium expressions The equilibrium constant of the second equation is the square root of the first. Question 2: When 82.9 grams of H 2 O (g) and 400. grams of Cl 2 O (g) are placed in a 190000 mL reaction vessel at 514 °C and allowed to come to equilibrium the mixture contains 3.86 mol of HOCl (g). How many mol / L of H 2 O (g) reacted? Solution: Determine the equilibrium concentration of HOCl Employ the stoichiometric relationship of HOCl and H 2 O to determine the concentration of H 2 O that reacted. Equilibrium Concentrations Question 3: Calculate the value of the equilibrium constant (K C ) for the reaction displayed if the following concentration data were measured at 100. °K: [H 2 ] = 5.24 M; [D 2 ] = 5.24 M; [HD] = 7.26 M.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Solution: The equilibrium constant is equal to the concentrations of the products over the reactants at equilibrium. Each compound is also raised to the power of its stoichiometric coefficient. Question 4: The initial pressures for the compounds involved in the reaction displayed were determined to be P(H 2 ) = 5.90 atm; P(I 2 ) = 7.60 atm; P(HI) = 0 atm. Calculate the value of the equilibrium constant (K p ) at 427°C, if the equilibrium pressure of H 2 was 0.782. Solution: Determine how many atm of H 2 reacted by subtracting the equilibrium pressure from the initial pressure. 5.90 atm - 0.782 atm = 5.12 atm reacted. If 5.12 atm of H 2 reacted, then 5.12 atm of I 2 reacted. Determine the equilibrium pressure of I 2 by subtracting the atm reacted from the initial pressure. 7.60 atm - 5.12 atm = 2.48 atm. From the stoichiometric relationship determine how many atm of HI are formed when 5.12 atm of H 2 react. The partial pressures of the three gases at equilibrium are P(HI) = 10.24 atm; P(I 2 ) = 2.48 atm; P(H 2 ) = 0.782 atm.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}