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Chemical Equilibrium: Shifting Concentrations and Factors Affecting Reactions, Slides of Chemistry

Anand Agricultural UniversityChemistry

The concept of chemical equilibrium, focusing on how altering conditions can shift equilibrium concentrations in a reaction. It covers the equilibrium state, the role of the equilibrium constant k, and how temperature, pressure, and catalysts impact equilibrium mixtures. It also discusses the link between equilibrium and kinetics.

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Chapter 13, Chemical Equilibrium
You may have gotten the impression that when 2
reactants mix, the ensuing rxn goes to completion.
In other words, reactants are converted completely
to products. We will now learn that is often not the
case, at least not in a rigorous, quantitative sense.
We will now learn:
1. to quantitatively describe how far a given
reaction proceeds toward completion.
2. how altering conditions can shift equilibrium
concentrations in a reaction.
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Partial preview of the text

Download Chemical Equilibrium: Shifting Concentrations and Factors Affecting Reactions and more Slides Chemistry in PDF only on Docsity!

Chapter 13, Chemical Equilibrium

You may have gotten the impression that when 2

reactants mix, the ensuing rxn goes to completion.

In other words, reactants are converted completely

to products. We will now learn that is often not the

case, at least not in a rigorous, quantitative sense.

We will now learn:

  1. to quantitatively describe how far a given reaction proceeds toward completion.
  2. how altering conditions can shift equilibrium concentrations in a reaction.

I. The Equilibrium State

A. We will start by looking at a rxn. that clearly

does not go to completion:

N O 2 4 (g) W 2 NO 2 (g) colorless reddish-brown Aside on terms:

  1. “Stuff” on the left is reactant , on the right is product.
  2. The rxn. toward the right (forming NO ) is the 2 forward rxn. The leftward (forming N O ) is the 2 4 backward rxn.

We can study this reaction by introducing a pure sample of either N O or NO into a reaction vessel. 2 4 2

  1. What happens at the leveled off point? a) Is the system frozen in place? b) Are the reactions (forward and backward) still occurring, but at equal rates?

Aside: At the leveled off point we say that we have reached equilibrium. Do we ever really reach it, or are we approaching it?

B. At equilibrium, the forward and backward rates

of the reaction are equal. (See Fig. 13.2)

  1. Based on your work in Chap. 12, why is the rate of the forward rxn. decreasing as the rxn. proceeds?
  1. Likewise, why is the rate of the reverse rxn. increasing as the rxn. proceeds?

C. What would happen if you started with different

initial [N O ] or [NO ]? (See Table 13.1) 2 4 2

  1. [N O ] & [NO ] 2 4 2 at equil. depend on [N O ] & [NO ] 2 4 0 2 0
  2. The ratio [NO ] /[N O ] is a constant!!! (Table 13.1) 2 2 4 2
  1. Kc is constant for a specific rxn. at a specific temp.
  2. For our rxn. with N O : 2 4 At 25EC K c = [NO ] 2 = 4.64 x 10 2! 3 [N O ] 2 4

B. Units of K c

  1. If K c is shown with units, they are determined by the concentration units and the number of terms in the K c expression. From above:

[NO ] 2 (M) 2 2 K c = (^) [N O ] 2 4 = (^) (M) = M

  1. Your text says units are generally omitted when using K c , but this depends on the branch of chemistry.

C. Equilibrium constant for the reverse rxn.

What is K c for: 2 NO 2 (g) W N O 2 4 (g)?

(The reverse rxn. equilibrium constant is called K c N.) Try Probs. 13.1-.3, pp. 496-7. Key Concept Prob. 13.

B. Recall that there is a direct effect of temperature

on P: PV = n RT

  1. Because of this, the numerical values of K p and Kc are not likely to be equal.
  2. However, you can determine the relationship between the 2 K s: K p = Kc x (RT) ( c + d ) - ( a + b )

See text for derivation. (Can you do the derivation? Comment on derivations. )

On your own, Prob. 13.5 & 6.

C. Biologists alert!!!

  1. If you have an interest in respiratory physiology, you might use these constants. O levels (regarding 2 hemoglobin binding, etc. ) are usually expressed as partial pressures. (Recovery room?)
  2. Hemoglobin ~ half-saturated w/ PO at 26 mm Hg. 2

So far we have confined our comments to one phase systems. Heterogeneous equilibria are also interesting.

  1. A Kc expression for this rxn. would be:

[CaO] [CO ] 2 “ K c ” = [CaCO ] 3

  1. However, your text notes that as CaO and CaCO are 3 solids, so [CaO] and [CaCO ] can’t change. (Think 3 collisions.) We can factor out these constant terms:

[CaO] “ K c ” = [CO ] x 2 â[CaCO ] 3 ê

[CaCO ] 3 “ K c ” x â [CaO] ê = [CO ] 2

Because [CaO] and [CaCO ] are constant, we can 3 combine this term with the “ K c ” term:

[CaCO ] 3 “ K c ” x â [CaO] ê = Kc

This is because: a constant × a constant = a constant.

  1. Finally: K c = [CO ] 2
  2. Note that there may be additional reasons for leaving solid or pure liquid component concentration terms out of equilibrium constant expressions. Do Prob. 13.7, p. 502.

If we go to the lab we can measure Kc. People have done this. K c = 2.4 x 10 at 500 K. 47

What does this mean re. [H O], [H ], [O ] at equil? 2 2 2

Look at K c expression above to figure this out. If [H O] = 5 M, what would the [H ] x [O ] be? 2 2 2 2

Very small. Essentially all of the material in this system is present in the product, H O. 2

  1. Another example: H2(g) + I2(g) W 2 HI(g)

[HI]^2 K c = [H ] [I ] 2 2 At 700 K, Kc = 57.

If we have an equilibrium condition where both [H ] 2 and [I ] are 0.10 M, what is [HI]? 2

Solve for [HI]: [HI] = ( K c x [H ] x [I ]) 2 2 1/

round [HI] = (57.0 x 0.10 x 0.10) 1/2 = 0.75498 ˆ 0.

Here, a significant portion of material is present in both the reactant and product components.

Problem 13.8, p.503.

  1. Plug values for any set of [H ], [I ], and [HI] into 2 2 Qc. Then, compare value of Q c to value of Kc. Predict direction the system goes (towards products or reactants)

For ex., if [H ] = 0.07 M, [I ] = 0.2M & [HI] = 3.0 M, 2 2 will this go toward product (HI) or reactants (H & I )? 2 2 H 2(g) + I2(g) W 2 HI(g) [3.0 M]^2 Q c = [0.07 M] [0.2 M] = 9.0 M /0.014 M = 643 2 2

Because Q c is larger than Kc , & the system must go toward the appropriate equilibrium ratios, [HI] must get smaller and [H ] & [I ] bigger. That is, the rxn. 2 2 will go toward the reactants H & I. 2 2

  1. Summary: If Qc < Kc , rxn. goes toward products. If Q (^) c > Kc , rxn. goes toward reactants. If Q (^) c = Kc , rxn. is already at equilibrium.

See Fig. 13.5, if it helps. Do prob. 13.9, p. 505.

Do Key Concept Prob. 13.