Question 14 Determine the equilibrium constant for the above reaction. Not yet a

Question

Question 14 Determine the equilibrium constant for the above reaction. Not yet answered 0.1 mol of Ca(OH)2 is placed in 2Lof water and stirred. At equilibrium, at a certain temperature, 0.0314 mol of Ca(oH)2 dissolves. What is Keq? (Assume, the volume of the solution is the same as the initial volume of water.) Marked out of 1.00 Answer: Flag question Question 15 consider the reaction: 3A(g) B(s) 5C 2D(g) K 0.42 MT 1 Not yet answered [A] 0.75 M and ID] 0.26 M. Some B and C are present. What is Q? Marked out of 1.00 Flag question An Help: Q and K

Explanation / Answer

Given :

Amount of Ca(OH)2 = 0.1 mol

Volume = 2 L

Amount of Ca(OH)2 dissolved = 0.0314 mol

Reaction in water:

Ca(OH)2 (s) -- > Ca2+ (aq) + 2 OH- (aq)

Equilibrium constant = [Ca2+][OH-]2

(solid phase doesnot take part in the equilibrium constant )

Calculation of [Ca2+] and [OH-]

Amount of Ca(OH)2 dissolved = 0.0314 mol

Moles of Ca2+

= Number of moles of Ca(OH)2 X 1 mol Ca2+ / 1 mol Ca(OH)2

= 0.0314 mol Ca(OH)2 x 1 mol Ca2+ / 1 mol Ca(OH)2

= 0.0314 mol Ca2+

[Ca2+] = mol Ca2+ / volume in L

= 0.0314 mol / 2.0 L

=0.0157 M

Moles of OH-

= Moles of Ca(OH)2 X 2 mols of OH- / 1 mol Ca(OH)2

= 0.0314 moles of Ca(OH)2 x 2 mol OH- / 1 mol Ca(OH)2

= 0.0628 mol OH-

[OH-]= 0.0628 mol OH- / 2.0 L

=0.0314 mol OH-

Lets plug these value in equilibrium constant expression.

Keq = [Ca2+][OH-]2

= (0.0157) ( 0.0314)2

= 1.55 x 10-5

Q. 2

Solution :

Here Q is the reaction quotient of the reaction when reaction is not at equilibrium

Q = [D]2/ [A]3 …( solid phase does not take part )

Lets plug the concentration of A and D

Q = (0.26)2 / ( 0.75)3

= 0.160

Q for this reaction is 0.160

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.