18.3 quiz

Question 1

what happens to the amounts of reactants and products after a reaction has reached chemical equilibrium?

Answer 1

both the forward and reverse reactions continue, but no net change occurs in the concentrations cuz the rates are equal

(easier answer: The amounts of reactants and products stop changing because the reactions balance each other out)

Question 2

what are 3 stresses that can upset the equilibrium of a chemical system?

Answer 2

change in:
- temp
- concentration (of products/reactants
- pressure

Question 3

what does the value of the equilibrium constant tell you about the amounts of reactants and products present at equilibrium?

Answer 3

it tells you whether reactants or products are more common at equilibrium

• If K is large (>1), there are more products than reactants
• If K is small (<1), there are more reactants than products
• If K ≈ 1, there are similar amounts of reactants and products

Question 4

can a change in pressure shift the equilibrium position in every reversible reaction? explain

Answer 4

no!
pressure only changes equilibrium if there are gases with unequal moles (different amounts) on each side

Question 5

how can you use a balanced chemical equation to write an equilibrium constant expression?

Answer 5

K = [products] / [reactants]

the coefficients are exponents

Question 6

which of the following equilibrium constants indicates a reaction in which the amount of product is much larger than the amount of reactant at equilibrium? explain

a. Keq = 1 x 10^8
b. Keq = 3 x 10^-6

Answer 6

a. this is a large K (greater than 1), meaning there are 𝐦𝐨𝐫𝐞 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 than reactants
b. this is a small K (less than 1), meaning there are 𝐦𝐨𝐫𝐞 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬 than products

Question 7

The equilibrium mixture for the reaction 2HI(g)⇌H2(g)+I2(g) contains 0.050mol H2. How many moles of I2 and HI are present at equilibrium (Keq=0.018)?

Answer 7

I2 is the same as H2, so
𝐈𝟐 = 𝟎.𝟎𝟓𝟎 𝐦𝐨𝐥

for HI - (0.050)(0.050) / 0.018 = 0.139
√0.139 = 0.373 mol
𝐇𝐈 = 𝟎.𝟑𝟕𝟑 mol

Question 8

how can understanding Le Chatelier’s principle help chemists increase the percent yield of a reversible chemical reaction?

Answer 8

Le Chatelier’s Principle helps chemists increase percent yield by changing conditions (like concentration, pressure, or temperature) to shift the reaction toward more products
(easier answer: if you push a reaction (by changing conditions), it pushes back to rebalance itself)

for example:
* increase reactants or decrease products to make more products

• increase pressure (for gases) to favor the side with fewer gas molecules
• lower temperature for exothermic reactions to produce more products

Question 9

how is the equilibrium position of this reaction affected by the following changes?
2SO2(g) + O2(g) +2SO3(g) + heat
a. lowering the temperature
b. decreasing the pressure
c removing oxygen
d. adding sulfur trioxide (SO3)

Answer 9

a. favors 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 (more SO3)
b. favors 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
c. favors 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
d. favors 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬

Question 10

C(s) + H₂O(g) + heat ⇌ CO(g) + H₂(g)
how is the equilibrium position affected by the following changes?
a. Increasing the pressure
b. Lowering the temperature
c. Removing H₂
d. Adding water vapor (H₂O)

Answer 10

a. favors 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
b. favors 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
c. favors 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬
d. favors 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬

Question 11

The reaction in which ammonia is formed is N2(g) + 3H2(g) + 2NH3(g). At equilibrium, a 1-L flask contains 0.15 mol H2, 0.25 mol N2, and 0.10 mol NH3. Calculate Keq for this reaction.

Answer 11

Keq = (0.10)^2 / (0.25x0.15^3) = 11.85 = 𝟏𝟐 𝐫𝐨𝐮𝐧𝐝𝐞𝐝

Question 12

referring back to qs 11, how are Keq forward & Keq reverse related?

Answer 12

Keq forward = 12
Keq reverse = 0.083 (1/12)
*literally the inverse

Question 13

suppose the following system reaches equilibrium.

N2(g) + O2(g) ↔ 2NO(g)

An analysis of the equilibrium mixture in a 1-L flask gives the following results: nitrogen, 0.50 mol; oxygen, 0.50 mol; nitrogen monoxide, 0.020 mol. Calculate Keq for the reaction.

Answer 13

Keq​ = 0.020^2 / (0.50)(0.50) =
0.0016 or 1.6 x 10^-3

​

Question 14

At 750°C, the reaction
H₂(g) + CO₂(g) ⇌ H₂O(g) + CO(g)
reaches equilibrium in a 1.0 L flask. Initially, there are 0.10 mol H₂ and 0.10 mol CO₂. At equilibrium, there are 0.047 mol H₂O and 0.047 mol CO. Calculate Keq for the reaction.

Answer 14

0.10 - 0.047 - 0.053
(0.047)(0.047) / (0.053)(0.053) = .786 =
= 𝟎.𝟕𝟗

Question 15

At a specific temperature, the equilibrium reaction 2 NO₂(g) ⇌ N₂O₄(g) has Keq = 5.6. In a 1.0 L container, the equilibrium amount of N₂O₄ is 0.66 mol. What is the equilibrium concentration of NO₂?

Answer 15

5.6 = 0.66
x^2 5.6 = 0.66
0.66 / 5.6
x^2 = 0.1179
√0.1179 = 𝟎.𝟑𝟒

Question 16

The decomposition of hydrogen iodide to hydrogen and iodine occurs by the reaction:
2 HI(g) ⇌ H₂(g) + I₂(g)
Hydrogen iodide is placed in a container at 450°C. An equilibrium mixture contains 0.50 moles of hydrogen iodide. The equilibrium constant is 0.020 for the reaction. What are the equilibrium concentrations of iodine and hydrogen iodide in the equilibrium mixture?

Answer 16

HI = 𝟎.𝟓𝟎𝐌 (given)
H2 & I2 = (0.020 x (0.50^2) = 0.005
√12.5 = 3.5 M

Question 17

Ta

Answer 17

Tantalum

Question 18

W

Answer 18

Tungsten

Question 19

Re

Answer 19

Rhenium

Question 20

Os

Answer 20

Osmium

Question 21

Ir

Answer 21

Iridium

Question 22

Pt

Answer 22

Platinum

Question 23

Au

Answer 23

Gold

Question 24

Hg

Answer 24

Mercury

Question 25

Tl

Answer 25

Thallium

Question 26

Pb

Answer 26

Lead