Class 2 Module 2 - Dr. Lori Vermeulen PDF

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Dr. Lori Vermeulen...

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Module 2 What is equilibrium and how do we describe it quantitatively? Many chemical reactions are reversible. When the rate of the forward reaction is equal to the rate of the reverse reaction, the system is at equilibrium. For example - A bathtub filling up with water. So, when the water is going in the bathtub and coming out of the bathtub. The rate coming in will be the rate coming out. So the things to notice is the level of the water will remain same because the rate of water going in will be equivalent to the rate of water going out. It is similar to the chemical reaction

Figure a is a hypothetical reaction where the red dots are the reactants and blue dots are the products and at the beginning the red dots are being converted to blue dots. As soon as it starts to form blue dots the reverse reaction starts happening and eventually get to a point where

the rate of the forward reaction is equal to the rate of the reverse reaction and that is what we call state of equilibrium. Other way to look at that is to look at the concentration of the reactants as shown in figure b. Initially we have all reactants no products after some time start to form some products, products start to form some reactants and when it gets to the state of equilibrium the concentrations of the products and reactants do not change.

aA

+

bB

⇄

cC

+

dD

• ⇄ ฀฀ℎ฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀ ฀฀฀฀฀฀ ฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀. • At equilibrium, the forward reaction and the reverse reaction are occurring at the same rate. • At equilibrium, the concentrations of the reactants and products are constant. • At equilibrium, it looks like nothing is happening, but really both the forward reaction and the reverse reaction are occurring. • The equilibrium constant, Kc (or Keq or Kp) defines the state of equilibrium and is a constant at constant temperature. • ⇄ ฀฀ℎ฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀ ฀฀฀฀฀฀ ฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀. • The equilibrium constant, Kc (or Keq) defines the state of equilibrium quantitatively. • ฀฀฀฀ = • • • •

[฀฀ ]฀฀[฀฀]฀฀

[฀฀]฀฀ [฀฀ ]฀฀

a, b, c, and d are the stoichiometric coefficients [ ] indicate concentration If H2O is the solvent, it does not appear in the ฀฀฀฀ expression. Solids do not appear in the ฀฀฀฀ expression.

• ⇄ ฀฀ℎ฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀ ฀฀฀฀฀฀ ฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀. • ฀฀฀฀ =

[฀฀ ]฀฀[฀฀]฀฀

[฀฀]฀฀ [฀฀ ]฀฀

• If ฀฀฀฀ is a really big number, the reaction is said to be productfavored or “the reaction lies to the right.” • If ฀฀฀฀ is a really small number, the reaction is said to be reactantfavored or “the reaction lies to the left.” • The value of ฀฀฀฀ depends upon the temperature. • If all of the reactants and products are gases, we can use pressure in place of concentration and calculate Kp instead of Kc. Example CH4(g) + H2O(g)

⇄

CO(g) + 3H2(g)

• Write the Kc expression and the Kp expression for this reaction. What is the relationship between Kc and Kp? [฀฀฀฀][฀฀2 ]3 4 ][฀฀2 ฀฀]

฀฀฀ ฀ =[CH

฀฀฀฀ = ฀฀฀฀฀฀

฀฀฀ ฀ = ฀฀฀฀฀฀ ·HP2 /P฀฀฀฀4 · PH2 ฀฀

฀฀฀฀ = � � ฀฀฀฀ ฀฀

฀฀฀

฀

{[฀฀฀฀]฀฀฀฀ · {[H2 ]R}3 } = [฀฀฀฀4 ]฀฀฀฀ · [H2 ฀฀] · RT

฀฀฀ ฀ = ฀฀฀ ฀ (฀฀฀฀)฀฀฀฀฀฀฀฀฀฀n

฀฀

Problem 2SO2(g) + O2(g) •

⇄

2SO3(g)

At equilibrium, •

The partial pressure of SO2 = 0.0018atm

•

The partial pressure of O2 = 0.0032 atm

•

The partial pressure of SO3 = 0.0166atm

•

What is the value of Kp?

2 ฀฀ ฀฀฀฀3

฀฀฀ ฀ = 2 ฀฀฀฀฀฀2 · ฀฀฀฀2

(0.0166)2 ฀฀฀ ฀ = (0.0018)2 ·(0.0032)

฀฀฀ ฀ = 2.7 × 104 Problem

2SO2(g) + O2(g) ⇄ 2SO3(g) • What is the value of Kp for the reverse reaction? 1 ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀ = ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀ 1 −5 = 4 = 3.7 × 10 2.7×10

Problem

SO2(g) +

1/2O2(g)

⇄ SO3(g)

• What is the value of Kp for the reaction written this way?...