Question

SHOW ALL WORK

1) The air pollutant NO is produced in automobile engines from the high-temperature reaction, N2 (g) + O2 (g) ⇌ 2 NO (g); Kc = 1.7 × 10-3 at 2300K . If the initial concentrations of N2 and O2 at 2300 K are both 1.4 M, what are the concentrations of NO, N2, and O2 when the reaction mixture reaches equilibrium? Show your work (20 points).

2) For each of the following equilibria, use LeChatelier’s principle to predict the direction of reaction when the volume is increased. (15 points)

(a) CO(g) + H2(g) ⇌ C(s) + H2O(g)

(b) 2H2(g) + O2 (g) ⇌ 2 H2O(g)

(c) Fe2O3(s) + 3 H2(g) ⇌ 2Fe(s) + 3 H2O(g)

3) Will the concentration of NO2 increase, decrease, or remain the same when the equilibrium of the exothermic reaction, NOCl(g) + NO2(g) ⇌ NO2Cl (g) + NO(g) is disturbed by the following changes? (15 points)

(a) adding NOCl

(b) adding NO

(c) Removing NO

(d) Adding argon

(e) addition of a catalyst

4) Calculate [H+] of a solution prepared by dissolving 4.25 grams of lithium hydroxide (24 g/mole) in water (18 g/mole) to give 350.0 mL of solution. (12 points) Kw = 1.0 × 10-14

5) Lactic acid (C3H6O3), which occurs in sour milk and foods such as sauerkraut, is a weak monoprotic acid. The [H+] of a 0.10 M solution of lactic acid is 4.57 × 10-3. What is the value of Ka for lactic acid? (16 points)

6) Write a chemical reaction for each of the following ions reacting with water. Demonstrate with this reaction whether they give neutral, acidic, or a basic solution. (16 points)

(a) F-

(c) NH4+

(d) K(H2O)6+

(e) SO32-

7) Calculate the percent dissociation of 0.10 M hydrazoic acid (HN3; Ka = 1.9 ×10-5) in the presence of 0.10 M HCl, and explain if this dissociation is more, the same, or less than if the HCl was not present. (16 points)

Answer 1

1) We write the reaction as

N₂ (g) + O₂ (g) --------> 2 NO (g)

The equilibrium constant for the reaction is

K_c = [NO]²/[N₂][O₂]

The equilibrium constant is constant at a particular temperature. the concentrations are supplied in moles/liter. We shall use the ICE table to calculate the equilibrium concentrations of all the species.

N₂ + O₂ --------> 2 NO

initial                                   1.4     1.4                  0

change                                 -x        -x                   2x

equilibrium                    (1.4-x) (1.4-x)              0+2x

We substitute the values in the equation for K_c.

K_c = (2x)²/(1.4-x)(1.4-x)

or, 1.7 x 10^-3 = 4x²/ (1.4 – x)²

Taking positive square root on both sides, we have,

0.0412 = 2x/1.4-x

or, x = 0.0282

Therefore, the equilibrium concentrations are

[N₂] = (1.4 – 0.0282) = 1.3718 M

[O₂] = (1.4 – 0.0282) = 1.3718 M

[NO] = (2 X 0.0282) = 0.0564 M (ans)

2a) When the volume of a system is increased at equilibrium, the system will adjust in such a way so as to nullify the effect of the change. Now, assuming ideal gas behavior, we know that pressure and volume are inversely related, i.e, increased volume leads to reduced pressure and viceversa. In the reaction,

CO (g) + H₂ (g) <======> C (s) + H₂O (g),

we see that the partial pressure of the gases is more on the reactant side (2 gaseous components); hence lower volume. So, increasing the volume of this system will lead to the side of lower partial pressure which is the product side. The forward reaction will be favored and we will obtain more products.

b) For the reaction, 2H₂ (g) + O₂ (g) <=======> 2H₂O (g), there is a reduction in partial pressure on going from the reactant to the product side. Increased volume will favour the side of lower partial pressure, i.e, the product side. More reaction will occur to afford the product.

c) Fe₂O₃ (s) + 3H₂ (g) <======> 2Fe (s) + 3H₂O (g)

In this reaction, we have equal number of moles of reactants and products (also partial pressures of the gaseous components will be the same). Increase in volume will have no effect on this system.

3) NOCl (g) + NO₂ (g) <=======> NO₂Cl (g) + NO (g)

The equilibrium constant is given as

K_c = [NO₂Cl][NO]/[NOCl][NO₂]

a) If more NOCl is added, the denominator of the above expression increases. To keep K_c constant (we assume K_c to be only temperature dependent), the numerator of the above expression must increase. This means the forward reaction will be favoured and we will obtain more products.

b) If more NO is added to the system at equilibrium, the numerator in the expression increases. To keep K_c constant, the denominator must also increase, i.e, the reverse reaction will be favoured. Formation of the products will be suppressed and we will have more of the reactants.

c) Removing NO will decrease the numerator. To keep K_c constant, the denominator must decrease. This is possible when the forward reaction is favoured,i.e, we shall have more products here.

d) Here, it is not clear if the volume of the system is held constant or not. If the volume is held constant, then addition of Ar (an inert gas) will have no effect on the system at equilibrium.

e) Catalyst has no effect on a system at equilibrium. It can, at best, accelerate or decelerate both the forward and reverse reactions.

4) We have the following dissociation : LiOH ---------> Li⁺ + OH^-

Now, LiOH is a strong base and will undergo complete dissociation. So, 1 mole of LiOH shall produce 1 mole each of Li⁺ and OH^-. Now, OH^- is basic, so we shall first calculate the pOH of the solution.

amount of LiOH taken = 4.25 g/ (24 g/mol) = 0.177 mole

The total volume of the solution is 350 mL.

So, the molar concentration of OH^- is (0.177 mole/350 mL).(1000 mL/1 L) = 0.5057 M

So, pOH = -log₁₀[OH^-] = -log₁₀(0.5057) =0.296

Now pH + pOH = 14

Therefore, pH of the solution is (14 – 0.296) = 13.704 (ans)

5) For a weak monoprotic acid, we have

HA <========> H⁺ + A^-

K_a = [H⁺][A^-]/[HA]

The [H⁺] is given as 4.57 x 10^-3. Since, the weak acid undergoes 1:1 molar dissociation (i.e, produces 1 mole of H⁺ and A^-), we obtain [A^-] as 4.57 x 10^-3. Also, [HA] = 0.10 M (given)

Hence, K_a = (4.57 X 10^-3)(4.57 X 10^-3)/(0.10) = 2.088 x 10^-4 (ans)

6a) F^- + H₂O --------> HF (weak acid) + OH^- (basic)

Solution will be basic due to hydroxide

c) NH₄+ + H₂O -----------> NH₄OH (weakly basic) + H⁺ (acidic)

Solution should be acidic.

d) K(H₂O)₆⁺ + H₂O ---------> KOH (strongly basic)

Solution should be basic.

e) SO₃^-2 + H₂O -----------> H₂SO₄ (strongly acidic)

Solution should be strongly acidic.