CHAPTER 14 BASIC HOMEWORK:
* Remember NOT to include liquids and/or solids in the expression.
Again tricky tricky no solids and/or liquids.
YOU GOT THIS KIDDOS
Consider the reaction: 3A(g) + B(s) ↔ 5C(s) + 2D(g).
In a 5.00 L container the reactants and products are at equilibrium. There are 0.44 mol A, 0.30 mol B, 1.43 mol C, and 0.47 mol D. What is the equilibrium constant?
Additional hint in feedback
Hint, remember to first write the equilibrium expression and convert the moles to concentrations.
In a 2.00 L container the reactants and products are at equilibrium. There are 0.47 mol A, 0.58 mol B, 1.84 mol C, and 0.27 mol D. What is the equilibrium constant?
3H2 = 4 – 3(0.86) = 1.42
2NH3 = 0 + 0.86 = 0.86
Kc = [0.86]^2/ [1.14][1.42]^3= 0.145
What is y, that is, how much hydrogen is used?
(Hint, use one of the methods used in the balance equation review problems to do the calculation. Enter amount as a positive quantity. An additional hint is given in the feedback.)
Additional hint: What is x? It is the amount of nitrogen used. What is the relationship between the amount of nitrogen used and the amount of hydrogen used?
What is z, that is, how much ammonia is produce?
What is the equilibrium constant, Keq?
(Remember to convert the number of moles to moles per liter or concentration in the equilibrium expression.)
What is the equilibrium expression for:
Ca(OH)2(s) ↔ Ca2+(aq) + 2OH–(aq)?
IF YOU’RE NOT A GAS YOU CANT SIT WITH US
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, 0.0224 mol of Ca(OH)2 dissolves. What is Keq? (Assume, the volume of the solution is the same as the initial volume of water.)
Consider the reaction: 3A(g) + B(s) ↔ 5C(s) + 2D(g) Keq = 0.42 M-1.
[A] = 0.73 M and [D] =0.22 M. Some B and C are present. What is Q?
Consider the reaction: 3A(g) + B(s) ↔ 5C(s) + 2D(g) Keq = 0.42 M-1
Same reaction and Keq, but this time:
[A] = 0.94 M and [D] =0.95 M. Some B and C are present. What is Q?
For the above value of Q
Q is large than Keq so to obtain equilibrium the reaction must move to the left so there is more reactant and less product.