Equilibria part iii
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Equilibria: Part III. Reviewing the rules. Things to remember. When reacting species (both the reactants and the products) are in their most condensed form (solids), the units are mol/L (molarity) and the constant is expressed as Kc. Things to remember.

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Equilibria: Part III

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Equilibria part iii

Equilibria: Part III

Reviewing the rules


Things to remember

Things to remember

  • When reacting species (both the reactants and the products) are in their most condensed form (solids), the units are

    • mol/L (molarity)

      and the constant is expressed as

    • Kc


Things to remember1

Things to remember

  • When species are in a gaseous phase

    state, the concentrations can be

    expressed in:

    • mol/L (Kc)

      OR

    • atm (Kp)


Things to remember2

Things to remember

  • The equilibrium constant is a dimensionless quantity.

    • MEANING: the equilibrium constant (Kp or Kc) has no units.


Things to remember3

Things to remember

  • Because the constant only applies to the reaction occurring at a specific temperature that does not change throughout the whole reaction, we MUST state the temperature and the balanced equation to which the equilibrium constant is referring.

    EXAMPLE: "CO(g) + 2H2(g)↔CH3OH(g) has a equilibrium constant of 10.5 at 220°C"


Things to remember4

Things to remember

5. If you know the equilibrium constant going in one direction (say from left to right), the equilibrium constant in the other direction (say right to left) is the inverse of that value.

ClNO2(g) + NO(g) → NO2(g) + ClNO(g)

Kc = 1.3 x 10^4 at 25° C

NO2(g) + ClNO(g) → ClNO2(g) + NO(g)

Kc = 1/(1.3 x 10^4 at 25° C)

or...

7.7 x 10^-5 at 25° C


Things to remember5

Things to remember

6. This is how you convert an Equilibrium constant from Kc to Kp:

Kp = Kc (0.0821 T)^Δn

T is the temperature during the reaction in Kelvin. Remember that Kelvin has exactly the same degree of difference between integers as celsius, but it does have a different starting point. Kelvin's zero is 273° less than celsius. For example, 2° C is equal to 275° K. 27° C is equal to 300° K.

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants


Things to remember6

Things to remember

  • Heoterogeneous equilibria refers to when a condensed solid or liquid interacts with a gas. To calculate the equilibrium constant, solve only for the gaseous state during such a multi-phase state reaction.


New rule

NEW RULE!!!

8. If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions.

Kc = (K'c) (K"c)


Let me elaborate

Let me elaborate...

1st Step Reaction

2nd Step Reaction

Overall

Reaction


To find the overall equilibrium of a multi step reaction

To find the overall equilibrium of a multi-step reaction


2 step reaction

2 step reaction

1st step reaction

N2(g) + O2(g) ↔ 2 NO(g) Kc1 = 2.3 x 10^-19

2nd step reaction

2 NO(g) + O2(g) ↔ 2 NO2(g) Kc2 = 3 x 10^6

Write the equilibrium equation for this multi-step reaction:

Kc = Kc1 x Kc2 = (2.3 x 10^-19)(3 x 10^6)

Kc= 7 x 10^-13


Equilibria part iii

B


Equilibria part iii

P


Equilibria part iii

1/[O2(g)]

D

WHY?

P


Putting it all together

Putting it all together

  • At the start of a reaction, there are 0.449 mol N2, 0.0221 mol of H2, and 0.000942 mol of NH3, in a 4.0-L reaction vessel at 375°C. If the equilibrium constant (Kc) for the reaction

    N2(g) + 3 H2 (g)→ 2 NH3 (g)

    Is 2.2 at this temperature, decide whether the system is at equilibrium. If not, predict which way the net reaction will proceed.


Step 1

Step 1

  • find the molarity of each species in the reaction:

    • molarity is mol/L

    • divide all mole values by 4 to find mol/L (because the total solution makes up 4 L).


Step 2

Step 2

  • Set up these initial values as a quotient expression (in the same manner as you would in finding the equilibrium constant)


Step 3

Step 3

  • Compare Qc to Kc (Kc = 2.2)

    • if Qc > Kc, then...

      • the initial concentrations of all the species are not in equilibrium. To reach equilibrium, the products must convert to reactants. The reactions will travel from right to left.

    • if Kc > Qc, then...

      • the initial concentrations of all the species are not in equilibrium. To reach equilibrium, the reactants must convert to products. The reaction will travel from left to right.

    • if Kc = Qc, then the reaction is in equilibrium.

      Nothing will happen. There will be no reaction.


Equilibria part iii

  • 2.9375 > 2.2

  • Qc > Kc

  • For the reaction to reach equilibrium, more products must converted into reactants

  • The reaction will move from right to left


Bellringer

BELLRINGER

f


Practice problem

PRACTICE PROBLEM

LET'S LOOK AT HOW TO SOLVE THIS...

g


Let s check this answer by substitution

LET'S CHECK THIS ANSWER BY SUBSTITUTION

  • WRITE OUT THE Kc EXPRESSION FOR THE 1ST STEP REACTION:

  • WRITE OUT THE Kc EXPRESSION FOR THE 2ND STEP REACTION:

d

d

d

A = 1 mol/L

B = 2 mol/L

C = 0.02 mol/L


Now let s change directions and add the coefficients

Now let's change directions and add the coefficients

  • 2nd STEP REACTION IN REVERSE:

  • 2nd STEP REACTION WITH COEFFICIENTS:

s

s

s

s

s

s

Therefore, in the future, we can just raise the Kc to a power equal to that of the stoichiometric coefficients belonging to the species.


Now change the direction and add coefficients to reaction 1

Now change the direction and add coefficients to Reaction 1

  • RAISE THE 1st STEP Kc TO THE POWER EQUAL TO THAT OF THE STOICHIOMETRIC COEFFICIENTS OF THE SPECIES:

Multiply the Kc' and Kc" to find the net Kc

Kc'Kc" = Kc = 0.25 x 10000 = 2500 (answer B)


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