14: Determine the equilibrium constant for the above reaction. 0.1 mol of Ca(OH)
ID: 1016838 • Letter: 1
Question
14: Determine the equilibrium constant for the above reaction.
0.1 mol of Ca(OH)2 is placed in 2 L of water and stirred. At equilibrium, at a certain temperature, 0.0250 mol of Ca(OH)2dissolves. What is Keq? (Assume, the volume of the solution is the same as the initial volume of water.)
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15: Consider the reaction: 3A(g) + B(s) 5C(s) + 2D(g) Keq = 0.42 M-1.
[A] = 0.68 M and [D] =0.26 M. Some B and C are present. What is Q?
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17: Consider the reaction: 3A(g) + B(s) 5C(s) + 2D(g) Keq = 0.42 M-1
Same reaction and Keq, but this time:
[A] = 0.86 M and [D] =0.96 M. Some B and C are present. What is Q?
Explanation / Answer
14. Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)
Equilibrium constant is the ratio of Concentration of product by the concentration reactant
Keq = [Ca2+][OH-]2
[Ca(OH)2]
We have to convert moles to molarity
Molarity = moles / volume
[Ca2+] = moles of Ca2+/2L of H2O = 0.025 mol /2 L
= 0.0125 mol/ L
2 X [OH-] = 2 X moles of OH- /2L of H2O
= 2X 0.025 mol/2L
= 2X 0.0125 mol/ L
= 0.025 mol/L
At equilibrium Ca(OH)2 not exit because it completely dissociate in to Ca2+ & OH- Ion
Since
Keq = [Ca2+][OH-]2
=0.0125M X (0.025)2M2
Keq =7.8125 x 10-6 or 0.78125 x 10-5 M3 M = mol/L
15. 3A(g) + B(s) 5C(s) + 2D(g) Keq = 0.42 M-1.
Q is reaction quotient,
Q = [C]5 [ D]2
[A]3 [B]
Given [A] = 0.68 M and [D] =0.26 M. Some B and C are present.
We didt know the concentration of B & c
Since Q is
Q = [ D]2
[A]3
= (0.26 M)2/ (0.68M)3
= 0.0676/0.3144
Q = 0.2150 M-1
Keq is 0.4 & Q is 0.2 what conculsion made for these two values
That is If K > Q, a reaction will proceed forward, converting reactants into products
16. 3A(g) + B(s) 5C(s) + 2D(g) Keq = 0.42 M-1.
Q is reaction quotient,
Q = [C]5 [ D]2
[A]3 [B]
Given [A] = 0.86 M and [D] =0.96 M. Some B and C are present.
We didt know the concentration of B & c
Since Q is
Q = [ D]2
[A]3
= (0.96 M)2/ (0.86M)3
= 0.9216/0.8847
Q = 1.0417M-1
Keq is 0.4 & Q is 0.1 what conculsion made for these two values
That is If K < Q, a reaction will proceed reverse direction, converting products into reactants
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