1040Online Dry Lab C - Online Dry Labs PDF

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COMPUTER LAB: GASEOUS EQUILIBRIA This online “dry” lab on “Gaseous Equilibria” is divided into two major parts: PART 1: Effect of Kc upon Changes in Concentrations PART 2: Effect of Kc upon Changes in Temperatures You need to go through the online lab (seven experiments) step-by-step and record all the necessary data (i.e. initial and equilibrium concentrations of N2 O4 and NO2 ) for the calculations of Kc and Q. The recorded data must be exact with precise numbers for the correct significant figures of Kc or Q. The equilibrium gaseous reaction studied in this experiment is: N 2O 4 (g) º 2 NO 2(g) (dinitrogen tetroxide) (nitrogen dioxide) (colorless) (red) The equilibrium constant expression for this reaction is:

Volume of reaction vessel =

PART 1 (A)

1L

Effect of Kc on Changes in Concentrations

At Equilibrium Conditions

EXPERIMENT #1

Initial at t = 0:

N2 O4 = 10 moles [N 2O 4] 0 = 10 M

NO2 = Ø moles [NO 2]0 = Ø

At Equilibrium:

[N 2O 4] = 9.416 M

[NO2 ] = 1.167 M

JCHU*1040

COMPUTER LAB: GASEOUS EQUILIBRIA

EXPERIMENT #2

Initial at t = 0: At Equilibrium:

N2 O4 = 1 moles [N 2O 4] 0 = 1 M [N 2O 4] = 0.827 M

Is Keq for this reaction the same as the Keq calculated in Experiment #1 ?

EXPERIMENT #3

Initial at t = 0: At Equilibrium:

N2 O4 = Ø moles [N 2O 4] 0 = Ø [N 2O 4] = 2.216 M

Compare this K eq to the K eq values calculated in Experiment #1 and #2.

(B)

PAGE - 2

NO2 = Ø moles [NO 2]0 = Ø [NO2 ] = 0.345 M

Yes

NO2 = 5 moles [NO 2]0 = 5 M [NO2 ] = 0.567 M

Similar

Under Non-Equilibrium Conditions When not at equilibrium, the reaction quotient (Q) is:

(1)

When the value of Q is smaller than K eq (i.e. Q < Keq ), the forward reaction will dominate and more NO2 (product) will be formed.

(2)

When the value of Q is larger than Keq (i.e. Q > Keq ), the reverse reaction will dominate and more N 2O 4 (reactant) will be formed. Q < K eq, the reaction shifts to product side. Q > K eq, the reaction shifts to reactant side.

JCHU*1040

COMPUTER LAB: GASEOUS EQUILIBRIA

EXPERIMENT #4

Initial at t = 0:

How will the [N 2O 4] change ? How will the [NO 2 ] change ?

At Equilibrium:

Conclusion:

EXPERIMENT #5

[N 2 O4 ]0 = 0.04 M

PAGE - 3

[NO2 ]0 = 0.06 M

decrease increase

[N 2O 4] = 0.034 M

[NO2 ] = 0.070 M

Since Q (0.09) < K (0.14), reaction shifts to product side. Q is less than K, so the reaction will make more products.  More NO 2 is produced, until equilibrium is reached.

Initial at t = 0:

How will the [N 2O 4] change ? How will the [NO 2 ] change ?

At Equilibrium:

[N 2 O4 ]0 = 4 M

[NO2 ]0 = 6 M

increase decrease

[N 2O 4] = 6.512 M

[NO2 ] = 0.975 M

Conclusion: Since Q (9) > K (0.146), reaction shifts to reactant side. Q is greater than K, so the reaction will make more reactants.  More N2 O4 is produced, until equilibrium is reached.

JCHU*1040

PART 2

COMPUTER LAB: GASEOUS EQUILIBRIA

PAGE - 4

Effect of Kc on Changes in Temperatures

(1)

At equilibrium, the chemical system appears static and the reactant and product concentrations no longer change. < This occurs not because the system is static, but because the forward and reverse reactions exactly balance each other (i.e. as much product is produced by the forward reaction as is consumed by the reversed reaction).

(2)

If either or both of the forward and reverse reaction rates change, then the concentrations of the reactants and products will change until a new balance is reached.

Influence of Temperature Changes on K c of Exo- / Endo-thermic Reaction C Increasing the temperature affects reaction in different manners. (1)

For the endothermic reaction (i.e. )H = r ve), increasing the temperature increases K c, more products are formed. or decreasing the temperature decreases Kc , more reactants are formed.

(2)

For the exothermic reaction (i.e. )H = È ve), increasing the temperature decreases Kc , more reactants are formed. or decreasing the temperature increases K c, more products are formed.

EXAMPLE

Consider the following endothermic reaction: 2 SO3 (g) + heat ø 2 SO2 (g) + O2 (g)

Increasing Temperature

Decreasing Temperature

larger Kc

smaller Kc

Reaction shifts to product side.  More SO2 and O2 are produced.

Reaction shifts to reactant side.  More SO 3 is produced.

Kc increases as the numerator is bigger.

Kc decreases as the denominator is bigger.

JCHU*1040

COMPUTER LAB: GASEOUS EQUILIBRIA

EXPERIMENT #6 At 318 K Initial at T = 298 K:

At Equilibrium (T = 318 K):

Conclusion:

PAGE - 5

[N2 O4 ]0 = 0.275 M

[NO2 ]0 = 0.199 M

[N 2O 4] = 0.200 M

[NO2 ] = 0.350 M

At higher temperature, the equilibrium shifts to product side. K eq increases as the temperature increases.  More NO 2 is produced, until equilibrium is reached.

EXPERIMENT #7

At 278 K

Initial at T = 298 K: At Equilibrium (T = 278 K):

[N2 O4 ]0 = 0.275 M [N 2O 4] = 0.327 M

[NO2 ]0 = 0.199 M [NO2 ] = 0.095 M

Predict how [N 2 O 4 ] and [NO2 ] will change when the temperature is decreased from 298 K to 278 K.

Î Ï Conclusion:

Keq decreases as the temperature decreases. The equilibrium shifts to reactant side. More reactant (N2 O4 ) and less product (NO2 )

At lower temperature, the equilibrium shifts to reactant side.  More N 2O 4 is produced, until equilibrium is reached.

Is the reaction exothermic or endothermic ?

endothermic...