### CHAPTER 14 BASIC HOMEWORK:

### EQUILIBRIUM II

**Resources:**

Have you found videos, websites, or explanations that helped you understand this chapter? Let us know and we’ll add them to “Resources” part of this page for other students to use.

**Question 1**

What is the equilibrium expression for:

2HI(g) ↔ H_{2}(g) + I_{2}(g)?

**Explanation**

Keq= [Gas Products]/ [Gas Reactants]

Keq= [H2][I2]/[HI]^2

**Question 2**

Which of the following show correct proportions for the above reaction?

(For d and e, x is a variable. For example, x could equal 0.3, or 2, etc.)

**Explanation**

HI:H2:I2

2:1:1

any variable of that

**Question 3**

For the reaction in the previous problem, that is,

2HI(g) ↔ H_{2}(g) + I_{2}(g) K_{eq} = 0.016

Initially a container contains 0.33 M HI and no product. What is the equilibrium concentration of H_{2}?

**Explanation**

Hint, in solving the equation you could use the quadratic equation, but in this case, it is easier to take the square root of both side and solve the simple algebraic equation.

2HI <–> H2+I2

I 0.3M 0M 0M

C- 2x +x +x

E 0.3-2x x x

keq = 0.016= x(x)/(0.33-2x)^2

sqrt (0.016)= x/(0.33-2x)

0.12649(0.33-2x)=x

0.041742-0.25298x=x

0.041742=1.25298x

x = 0.033

**Question 4**

For the reaction:

2HBr(g) ↔ H_{2}(g) + Br_{2}(g)

Initially a container contains 0.64 M HBr and no product. What is the equilibrium concentration of HBr if the equilibrium concentration of H_{2} is 0.20 M? (Hint provided in feedback.)

**Explanation**

#### Hint–how much HBr is needed to produce the H_{2}? Subtract this amount from the original amount to find out how much is left.

_{2}(g) + Br

_{2}(g)

**Question 5**

For the reaction:

2HBr(g) ↔ H_{2}(g) + Br_{2}(g)

0.45 M Br_{2} is at equilibrium with a certain concentration of HBr and H_{2}. The concentration of bromine is increase by 0.2 M and the system establishes a new equilibrium. Which of the following statements are true?

All answers must be correct to get credit.

Concentration and Le Chatelier’s Principle

**Explanation**

2HBr(g) ↔ H_{2}(g) + Br_{2}(g)

I

C

E x y 0.45M

2HBr(g) ↔ H_{2}(g) + Br_{2}(g)

I x y 0.65

C +2a -a -a

E x+2a y-a 0.65-a

HBr increases

Br2 decreases

H2 decreases

**Question 6**

For the reaction:

N_{2}(g) + 3H_{2}(g) ↔ 2NH_{3}(g)

Nitrogen, hydrogen, and ammonia are at equilibrium at 10 atm. The pressure is increased to 50 atm by decreasing the volume. The system moves to establishes a new equilibrium. Which of the following statements are true?

Pressure and Le Chatelier’s Principle

**Explanation**

N_{2}(g) + 3H_{2}(g) ↔ 2NH_{3}(g)

10 atm = equilibrium

P increased by V decreasing (P=nRT/V) –> inversely proportional

When P increases/V decreases equilibrium prefers the side with the least amount of moles of gas

….on the left there are 4 moles of gas and on the right there are 2

so it will go to the right

meaning that the moles of NH3 will increase and moles of N2 and H2 will decrease

**Question 7**

For the reaction:

N_{2}(g) + 3H_{2}(g) ↔ 2NH_{3}(g)

Nitrogen, hydrogen, and ammonia are at equilibrium at 10 atm. The pressure is dropped to 2 atm by increasing the volume. The system moves to establishes a new equilibrium. Which of the following statements are true?

**Explanation**

N_{2}(g) + 3H_{2}(g) ↔ 2NH_{3}(g)

When pressure decreases it favors the side with the most moles of gas

So it will go the left meaning that N2 and H2 will increase whilst NH3 will decreases

**Question 8**

For the reaction:

2HBr(g) ↔ H_{2}(g) + Br_{2}(g)

Hydrogen bromide, hydrogen, and bromine are at equilibrium at 8 atm. The pressure is decreased to 3 atm by increasing the volume. The system moves to establishes a new equilibrium. Which of the following statements are true?

**Explanation**

2HBr(g) ↔ H_{2}(g) + Br_{2}(g)

By decreasing the pressure and increasing the volume the reaction would favor the side with most moles of gas, but in this case they have equal cases of gaseous moles so the reaction will stay in equilibrium.

Meaning that nothing will change!!!

**Question 9**

For the reaction:

CaCO_{3}(s) ↔ CaO(s) + CO_{2}(g) K_{p}= 1.16 at 800^{o}C

The system is at equilibrium. There are 2 mol of CaCO_{3}, 6 mol CaO, and the CO_{2} pressure is 1.16 atm at 800^{o}C. Which of the following produces more CO_{2} at the new equilibrium?

**Explanation**

CaCO_{3}(s) ↔ CaO(s) + CO_{2}(g) K_{p}= 1.16 at 800^{o}C

I

C

E 2 6 1.16

In order to produce more CO2 the only possibility is to decrease pressure as on the left there is only a solid which plays no role in the equilibrium.

**Question 10**

For the reaction:

N_{2}(g) + 3H_{2}(g) ↔ 2NH_{3}(g) ΔH = -92 kJ

Which of the following statements are true?

**Explanation**

N_{2}(g) + 3H_{2}(g) ↔ 2NH_{3}(g) ΔH = -92 kJ

ΔH = -92 kJ === exothermic

meaning that heat is a product

so in order to increase ammonia concentration the temperature must decrease

**Question 11**

Pick the correct statements for the reaction given below. (To the right means the new equilibrium has more product, to the left means the new equilibrium has more reactant.)

Si(CH3)4 (g) + 12CO2(g) <–> SiO2(s) + 16CO(g)

**Explanation**

YOU GOT THIS KIDDOS

increasing the GAS products moves the equilibrium to the reactants and vice versa

decreasing the GAS products moves the equilibrium to the products and the same goes for the reactants

**Question 12**

Pick the correct statements for the above reaction.

Select one or more:

**Explanation**

All must be correct to get full credit. Remember that changing the amount of pure liquids and solids has no effect (as long as some remains).

H20 is a liquid so genuinely has no effect on anything-also it isn’t even in the en above

And by expanding the volume aka decreasing the pressure it will go to the right because there 16 moles of gas in comparison to the 12 moles of gas on the right

**Question 13**

Pick the correct statements for the above reaction.

Select one or more:

**Explanation**

#### Hint–an inert gas is not part of the equilibrium expression and has no effect on the equilibrium concentrations

**Question 14**

For the reaction:

2HI(g) ↔ H_{2}(g) + I_{2}(g) K_{eq} = 0.016

Initially a container contains 0.60 M HI, 0.038 M H_{2}, and 0.15 M I_{2} at equilibrium. What is the new equilibrium concentration of H_{2}, if the H_{2}concentration is increased by 0.190 M?

**Explanation**

2HI(g) ↔ H_{2}(g) + I_{2}(g) K_{eq} = 0.016

I 0.6 0.228 0.15

C +2x -x -x

E 0.6+2x 0.228-x 0.15-x

keq = 0.016 = [H2][I2]/[HI]^2

0.016= (0.228-x)(0.15-x)/(0.6+2x)^2

0.016= 0.0342-0.378x+x^2/(0.36+2.4x+4x^2)

0.0576 + 0.0384x + 0.064x^2= 0.0342 – 0.378x +x^2

-0.936x^2+ 0.4164x-0.02844=0

x= -0.4164(+/-) sqrt(0.17339-4(-0.936)(-0.02844))/(2*-0.936)

x= -0.4164 (+/-) sqrt(0.0669)/(2* -0.936)

x= 0.084258

0.228- (0.084258)=0.1437