Equilibrium constant Kc for homogeneous systems Flashcards
What is the formula for working out Kc? Why are the SQUARE brackets important here?
- Kc = [C]^c[D]^d / [A]^a[B]^b
- Square brackets = important, shows that we are working with equilibrium concentrations.
- Products/ reactants.
Write the EXPRESSION for Kc for the reaction below:
- 2SO₂ (g) + O₂ (g) –> (reverisble reaction sign) 2SO₃
What would the units be too?
- Expression: ⌊SO₃⌉²/ ⌊SO₂⌋² x ⌊O₂⌋
- Units: mol⁻¹dm³
How do you workout the units of Kc? Use example of this equation below:
- 2SO₂ (g) + O₂ (g) –> (reverisble reaction sign) 2SO₃
- Expression: ⌊SO₃⌉²/ ⌊SO₂⌋² x ⌊O₂⌋
- Replace molecules by moldm⁻³ sign SO….
- ⌊moldm⁻³⌋ x ⌊moldm⁻³⌋/ ⌊moldm⁻³⌋ x ⌊moldm⁻³⌋ x ⌊moldm⁻³⌋
- cancel out. You are left with: 1/ ⌊moldm⁻³⌋, so reciprocal it: mol⁻¹dm³
What does Kc value of 1 indicate? If Kc is a large value, where does this say the equilibrium is shifted to? How about if its a small value (less than 1?)
- Indicates that the reaction is at equilibrium.
- Lare value shows equilibrium has shifted to the right (because large value means you have more products, less reactants.)
- Less than 1 shows equilibrium shifted to left (because small value means you have more reactants than products.)
If I have reaction:
A + B —> (reversible reaction sign) C+D, and I know the Kc value in forward reaction = x, what is the expression for Kc value in backward reaction/ direction?
- 1/x
Example Q (using ICE!)
1 mol of acid reacted with 0.33mol of alcohol. At equilibrium , 0.293 mol of esther is present. Calculate no of moles at equilibrium.
CH₂COOH + CH₃CH₂OH –> (reversible reaction sign) CH₃COOCH₂CH₃+ H₂O
I(nitial moles)
CH₂COOH = 1mol/ CH₃CH₂OH = 0.33mol. Products = 0 mol each (at start of reaction.)
C(hange in moles)
CH₃COOCH₂CH₃ –> + 0.293
H₂O –> +0.293 (1:1 ratio)
CH₃CH₂OH 0.33-0.293 (1:1 ratio)
CH₂COOH: 1 - 0.293 (1: 1 ratio)
From left to right: 0.707, 0.037, 0.293, 0.293
What does ICE stand for when were are doing Kc calculations?
I(nitial no of moles)
C(hange in moles)
M(oles at equilibrium.)
If Q doesn’t give you volume of vessel of reaction (to calculate the concentrations at equilibrium so you can calculate value of Kc), what do you do?
- Just use the no of moles (no need to try and find concentration.) This is because the volume would just cancel out! This is the case if you have same powers on top and bottom for instance:
- A² x B²/ C² x D²
What type of systems do we calculate the equilibrium constant for? What does this mean?
- Homogenous systems.
- All of the molecules are in the same state.
If the forward reaction is endothermic and we increase temp what effect will this have on Kc value?Where does equilibrium position shift to?
- Increase Kc value
- Equilibrium positio shifts to right.
If the forward reaction is exothermic and we increase temp what effect will this have on Kc value?Where does equilibrium position shift to?
- Kc will decrease
- Equilibrium position shifts to left
Example
Reaction took place, equilibrium was reached. At equilibrium, conc of ethanoic acid = 1.8moldm-3 and ethanol = 3.2moldm-3. Value of Kc = 3.5 at 25 degrees celcius. Calculate conc of products at equilibrium.
Equation: CH₃COOH (l) + + C₂H₅OH(l) –> CH₃COOC₂H₅ (l) + H₂O (l)
CH₃COOC₂H₅ = 4.49moldm-3
H2O = 4.49 moldm-3
What is the only factor that changes Kc? What is Kc NOT AFFECTED by?
- Temperature
- Not affected by: changes in concentration/ addition of catalyst.
This is a reaction:
A + B —> (reversible reaction sign) C+ D.
This reaction has Kc value of 3.2 at 25 degrees, if i change the temp what will happen to the Kc value?
- The Kc value will change, because Kc is a temperature-dependent constant.
Addition of catalyst will not change Kc value, but what DOES addition of catclyst do?
- Speeds up forward/ reverse reaction at the same rate.
- Speeds up rate at which equilibrium is reached.
