Chemical equilibria, Le Chatelier's principle and Kc (3.1.6)

Question 1

what happens in a reversible reaction?

Answer 1

the products automatically react to reform the reactants

Question 2

what are the characteristics of a reversible reaction at dynamic equilibrium?

Answer 2

the forward and backward reactions proceed at equal rates so the concentrations of reactants and products remain constant

Question 3

what is needed for a reaction to be reversible and reach dynamic equilibrium?

Answer 3

a closed system

Question 4

what is Le Chatelier’s Principle?

Answer 4

if one or more of the factors that affect the position of equilibrium are altered, then the equilibrium shifts to reduce (oppose) the effects of the change

Question 5

what will be the effect of an increase in temperature on the position of equilibrium?

Answer 5

the position of equilibrium will shift to the endothermic side to oppose the increase in temperature

Question 6

what will be the effect of an increase in pressure on the position of equilibrium?

Answer 6

the position of equilibrium will shift to the side with fewer moles of gas to oppose the increase in pressure

Question 7

what will be the effect of an increase in concentration of products on the position of equilibrium?

Answer 7

the position of equilibrium will shift to the right (forwards reaction) to oppose the increase in concentration by using up the extra products

Question 8

what is the effect of a catalyst on position of equilibrium?

Answer 8

they have no effect because they increase the rate of the forwards and backwards reactions equally

Question 9

why might a compromise temperature and pressure be used for a reversible reaction in an industrial process?

Answer 9

because the temperature and pressure must give a balance between reasonable yield (based on position of equilibrium), acceptable rate and cost (high temperature increases cost as more energy is used/high pressure increases cost as the walls of the reaction vessels must be ticker so use more materials)

Question 10

what is an example of a reversible reaction in an industrial process which uses a compromise temperature and pressure?

Answer 10

the Haber process - 450 degrees, 200 atm, iron catalyst

Question 11

how is the equilibrium constant represented?

Answer 11

Kc

Question 12

where is the equilibrium constant deduced from?

Answer 12

the equation for a reversible reaction

Question 13

what does [X] represent?

Answer 13

the concentration in mol dm^-3 of a species (X) involved in the expression for Kc

Question 14

which things don’t affect the value of the equilibrium constant?

Answer 14

changes in concentration or addition of a catalyst

Question 15

how do you construct an expression for Kc for a homogeneous system in equilibrium?

Answer 15

• Kc = [products]/[reactants]
• balancing numbers in equation=powers

Question 16

what would the expression for Kc be for the following balanced equation:
N2 (g) + 3H2 (g) -><- 2NH3 (g)

Answer 16

Kc = [NH3]^2/[N2] [H2]^3

Question 17

how do you work out the units for the equilibrium constant?

Answer 17

1. add the indices on each level of the Kc expression
2. find the overall indice by subtracting the bottom overall indice from the top one
3. to get the unit do (mol dm^-3)^indice

Question 18

what would the unit for Kc be for the following expression:
Kc = [K]^3 [L]^2/ [I] [J]^2

Answer 18

1. []^5/ []^3
2. []^2
3. (mol dm^-3)^2 = mol^2 dm^-6

Question 19

how do you calculate a value for Kc from the equilibrium concentrations for a homogeneous system at a constant temperature?

Answer 19

1. substitute the values into the Kc expression
2. pay attention to the powers
3. calculate value and units

Question 20

how do you calculate a value for Kc when given general information about a reaction?

Answer 20

1. set up your work by writing out moles at start, change in moles, moles at equilibrium and concentration at equilibrium, and creating a column for each reactant and product
2. fill in what you already know from the question
3. fill in the rest of the information based on what you already know (change in moles will be based on molar ratios)
4. work out the concentration at equilibrium
5. write the Kc expression, substitue the values in, calculate and find the units

Question 21

what would the value for Kc and units be for the following reaction:
2.7 mol of CH4, 1.9 mol of H2O and 1.0 mol of H2 are mixed and left in a 1dm^3 sealed vessel until the reaction reaches equilibrium. At equilibrium there is 0.7 mol of CO2. CH4 + 2H2O -><- CO2 + 4H2

Answer 21

CH4 H2O CO2 H2
moles at start 2.7 1.9 0.0 1.0
change in moles -0.7 -1.4 +0.7 +2.8
moles at equilibrium 2.0 0.5 0.7 3.8
concentration at equilibrium 2.0 0.5 0.7 3.8

Kc = [CO2] [H2]^4/ [CH4] [H2O]^2
Kc = (0.7) x (3.8)^4/ (2.0) x (0.5)^2 = 292 mol^2 dm^-6

Question 22

deduce the value for Kc for following reaction - CO2 + 4H2 -><- CH4 + 2H2O.

Answer 22

1/292 = 0.00342 mol^-2 dm^6
when it asks you to deduce Kc for the reverse reaction of one you have already calculated it for, do 1/previous Kc value

Question 23

what are the qualitative effects of changes in temperature on the value of Kc?

Answer 23

• if change in temperature causes position of equilibrium to shift to right ie. higher concentration of products, Kc value will increase