Gen Chem II Chapter 15 Jeopardy

Kp & Kc Problems

100

Consider the following reaction:

2NOBr_(g) = 2NO_(g) + Br_2(g)

If [NOBr] = 0.46 M

[NO] = 0.1 M

[Br₂] = 0.3 M

Assuming this occurs at 20^oC, determine K_c & K_p for this reaction(R = 0.08206 (L*atm)/(mol*K)) (K_p = K_c(RT)^Δn)

K_c = ([NO]²[Br₂])/([NOBr]²)

= ((0.1² M)(0.3 M))/(0.46² M)

= 0.0142 M

K_p = K_c(RT)^Δn 

Δn = 3 - 2 = 1

T = 20+273

K_p = (0.0142 M)*((0.08206 (L*atm)/(mol*K))*(293 K))¹

= 0.341 atm

200

Consider the following equilibrium reaction

CO_(g) + H₂O_(g) = CO_2(g) + H_2(g)

Given that K_c = 1.00 and this reaction occurs at 1100 K, the reaction vessel has the following quantities:

CO = 1.00 mol

H₂O = 1.00 mol

CO₂ = 2.00 mol

H₂ = 2.00 mol

V = 1.00 L

Which direction will the equilibrium shift to?

Q = ([CO₂][H₂])/([CO][H₂O])

= ([2.00][2.00])/([1.00][1.00])

= 4.00

K_c < Q

Reaction equilibrium will shift left 

ICE Equation

100

Consider the following Generalized Reaction:

2X_(g) = 3Y_(g) + 4Z_(g)

0.500 mol of X is placed within an enclosed cylinder that has a volume of 0.750 L. When equilibrium is achieved, the amount of X is 0.350 mol. Determine K_c for this reaction.

#     2X       3Y       4Z

I    0.500     0         0

C    ?            ?        ?

E    0.350     ?        ?

Equilibrium amount = Initial amount + Change amount

Change amount = 0.350 - 0.500

= -0.150

Change in Product = -(Coeff. Product/Coeff. Reactant)(Change Reactant)

Change Y = -(3/2)(-0.150) = 0.225 mol

Change Z = -(4/2)(-0.150) = 0.300 mol

V container = 0.750 L

M X = 0.350/0.750 = 0.467 M

M Y = 0.225/0.750 = 0.300 M

M Z = 0.300/0.750 = 0.400 M

K_c = ((0.300³)(0.400⁴))/(0.467²)

= 3.17*10^-3 

200

For the following equilibrium reaction:

H₂O_(l) + CH₃COOH_(aq) = H₃O_(aq) + CH₃COO^-

If K_a = 1.8*10^-5 & [CH₃COOH] = 0.150M, determine the concentration of CH₃COO^-.

Rxn   CH₃COOH      CH₃COO^-      H₃O⁺

I          0.150 M       0 M             0M

C        -x                 +x               +x

E        0.150-x           x                 x

1.8*10^-5 = (x²)/(0.150-x)

x² = -(1.8*10^-5)x + (1.8*10^-5)(0.150)

x = 

Quadratic method (don't recommend)

0 = x² + (1.8*10^-5)x + (1.8*10^-5)(0.150)

x= (-b +- (b² - 4ac)^-2)/2a

= (-(1.8*10^-5) +- ((1.8*10^-5)² - 4(1.8*10^-5)))/2

Pick - sign for answer (answer must be positive)

x = 2.7 * 10^-5 M

(CHECK for accuracy)

Chemical Equilibrium

100

For the following dissolution:

Ag₂CrO_4(s) = 2Ag⁺_(aq) + CrO₄^2-_(aq)

Write the expression for K_sp for this reaction

K_sp = [Ag⁺]²[CrO₄^2-]

200

For the following Equilibrium Reaction:

2H₂SO_4(aq) + HNO_3(aq) = NO₂⁺_(aq) + H₃O⁺_(aq) + 2HSO₄^-_(aq)

Write an expression that will represent the Equilibrium constant for this Reaction

K = ([NO₂⁺][H₃O⁺][HSO₄^-]²)/([H₂SO₄]²[HNO₃])

300

Consider the following generalized equilibrium reaction:

aA + bB = cC + dD + eE

Can equilibrium be established if only B, C, & D are in the reaction vessel?

No, equilibrium cannot be established from either side. A would be needed for the reactants side, & E would be needed for the products side.

400

Calcium Oxalate Monohydrate Ca(O₂CCO₂)*H₂O is a salt that is sparingly soluble in water. It's solubility in water is 7.36*10^-4 g/100mL. Calculate it's K_sp. (Assume water's density is 1 g/mL

Ca(C₂O₄)*H₂O_(s) = Ca²⁺_(aq) + C₂O₄^2-_(aq) + H₂O_(l)

K_sp = [Ca²⁺][C₂O₄^2-]

MM C₂O₄^2- = 146.1 g/mol

S = ((7.36*10^-4 g )/(100 mL)) * (1000 mL/ 1 L) = 5.04*10^-5 mol/L or M

Since 1 mol of Calcium Oxalate Monohydrate produces 1 mol of the Calcium ion and 1 mol of the Oxalate ion, we can perform the following substitution on the K_sp expression

K_sp = [5.04*10^-5]²

= 2.54*10^-9

Le Chatlier's Principle

100

Consider the following Equilibrium Reaction:

A + 2B = C + D

If [A] is doubled, how will the equilibrium shift?

It will shift towards the products

200

Consider the following equilibrium reaction, which is occurring in a sealed vessel:

2NOCl_(g) = 2NO_(g) + Cl_2(g)

How will the reaction equilibrium shift if the pressure of the vessel is decreased?

The equilibrium will shift to the right (products)

Acid Base Equilibria

100

Determine if the following compounds are strong acids, strong bases, weak acids, or weak bases:

1. NH₃                                                    2. CH₃COOH

3. H₂SO₄                                               4. HF

5. HBr                                       6. Mg(OH)₂

7. LiOH                                      8. Ca(OH)₂

9. HNO₃                                               10. H₃PO₄

1. Weak Base                               2. Weak Acid

3. Strong Acid                              4. Weak Acid

5. Strong Acid                              6. Weak Base

7. Strong Base                             8. Strong Base

9. Strong Acid                             10. Weak Acid

200

Identify the Bronsted-Lowry Acids & Bases for the following reactions:

1. H₂SO_4(aq) +H₂O_(l) -> H₃O⁺ + HSO₄^-_(aq)

2. CH₃COOH_(aq) + NH_3(aq) -> NH₄⁺_(aq) + CH₃COO^-_(aq)

3. H₂PtCl_6(aq) + NaHCO_3(aq) -> NaHPtCl_6(aq) + H₂CO_3(aq)

1. Acid: H₂SO₄, Base: H₂O

2. Acid: CH₃COOH, Base: NH₃

3. Acid: H₂PtCl₆, Base: NaHCO₃

300

The pK_a of acetic acid (CH₃COOH) is 4.76 at 25 ^oC. For glacial acetic acid, which has a concentration of 17.0 M, determine [H⁺].

K_a = 10^-4.76 = 1.74*10^-5

CH₃COOH_(aq) = H⁺_(aq) + CH₃COO^-_(aq)

1.74*10^-5 = (x²)/(17.0 M)

x = (17.0*1.74*10^-5)^(1/2)

= 1.72*10^-2 M

400

An aqueous solution has a [H⁺] of 2.43*10^-3 M . Determine the pOH for this solution.

K_w = [H⁺][OH^-]

1.0*10^-14 M = [2.43*10^-3 M][OH^-]

[OH^-] = (1.0*10^-14)/(2.43*10^-3)

= 4.12*10^-12

pOH = -log(4.12*10^-12)

= 11.4