Name: Instructor: Partner(s): Date: CHAPTER 14 Concept Explorations 14.17. Chemi

Question

Name: Instructor: Partner(s): Date: CHAPTER 14 Concept Explorations 14.17. Chemical EquilibriumI Part 1: You run the chemical reaction C(aq) + D(aq) 2E(aq) at 25°C The equilibrium constant Ke for the reaction at this temperature is 2.0. a. Write the equilibrium-constant expression for the reaction. b. Can you come up with some possible concentrations of C, D, and E that you might observe when the reaction has reached equilibrium at 25°C? What are these values? A student says that only a very limited number of concentrations for C, D, and E are possible at equilibrium. Is this true? State why you think this is true or is not true. c. d. If you start with 1.0 Mconcentrations of both C and D and allow the reaction to come to equilibrium, would you expect the concentration of C to have decreased to zero? If not, what would you expect for the concentration of C? (An approximate value is fine.) Part 2: Consider the reaction A(aq) + B(a) F(aq) + G(aq), whose equilibrium constant is 1.0 × 10-5 at 20°C. For each of the situations described below, indicate whether any reaction occurs. If reaction does occur, then indicate the direction of that reaction and describe how the concentrations of A, B, F, and G change during this reaction. a. A(a) and B(a) are mixed together in a container.

Explanation / Answer

PART I

a)

the K expressoin

K = products^p / reactants ^r

then

K = [E]^2 / ([C][D])

b)

if K = 2

and you Kc; you need at least 1 value initially or values in equilibirum. No dat aof concnetration is given, therefore we can't assume any value. You may also include stoichiometric ratios of the feed

c)

this is true, since Kc must be always equal to "2"

thjen

K = [E]^2 / ([C][D]) turns to

2 = [E]^2 / ([C][D])

If you set C and D, then the vlaue of E is fixed. Note that in reality there are infinite number os possibilities.

d)

[C] = 1

[D] = 1

[E]= 0

would oyu expect C to become near to zero?

NO, since K is not >>> than 1

its approx 1 so, in magnitude, expect almost 50% products 50% reactants...

in equilibrium

K = [E]^2 / ([C][D])

2 = (0+2x)^2 / (1-x)(1-x)

solve for x

sqrt(2) = (2x) /(1-x)

1.41(1-x) = 2x

1.41 - 1.41x = 2x

3.41x = 1.41

x = 1.41/3.41 = 0.41348

[C] = 1-x = 1-0.41348 = 0.58652

which is 58.6% of what we talked about (near 50%)

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.