Equilibrium
, The equilibrium constant Kc
The equilibrium law tells us the relative proportions of
reactants and products when a reaction is at equilibrium.
aA+bB ⇄ cC+dD The square brackets show
Kc= [C] ͨ [D] ͩ [products] . concentration in moldm¯³
[reactants]
[A] ͣ [B]ᵇ The indices are the numbers in front of
Equilibrium constant each substance in the chemical equation
C shows that its in terms of
concentration
If Kc has a value of 1 then the
Examples position of equilibrium is
H₂ (g) +I₂(g) ⇄ 2HI(g) halfway between the reactants
and products
The equilibrium constants are If Kc>1 then the reaction is
H₂ 0.140 moldm¯³ product favoured so there are
HI 0.320 moldm¯³ more products at equilibrium
I₂ 0.0400 moldm¯³ If Kc<1 then the reaction is
Kc= [HI(g)]² Kc= [0.320(g)]² .=18.3 reactant favoured so there are
[H₂(g)][I₂(g)] [0.14(g)][0.04(g)] more reactants at equilibrium
, Example from experimental results
N₂(g) + 3H₂(g) 2NH₃(g)
2.5mol of N₂ and 6.2 mol of H₂ were placed in a 5dm³ container & were left
to reach equilibrium. 2 mol of NH₃ was measured at equilibrium
Calculate the Kc of this reaction
Step 1: work out the equilibrium amounts of all equilibrium species using
MICE (Eq)
ratio 1 3 2 if this was a
Molecules N₂ H₂ NH₃ heterozygous equilibrium
only (g) & (aq)would be
Initial moles 2.5 6.2 0 included in working out
Change in moles -1.0 -3.0 +2.0
If value of v is unknown
Equilibrium moles 1.5 3.2 2.0
just use ‘v’ as this will
Equilibrium concentration 1.5/5=0.3 3.2/5=0.64 20/5=0.4 cancel out in the Kc
(c=n/v) expression
Step 2: write expression for Kc
Kc= [NH₃]² units= [moldm¯³]² .
[N₂][H₂]³ [moldm¯³] [moldm¯³]³
Kc= 0.4² .
0.3 x 0.64³ units=mol⁻²dm⁶
= 2.03
2.03>1 so product favoured
, The equilibrium constant Kc
The equilibrium law tells us the relative proportions of
reactants and products when a reaction is at equilibrium.
aA+bB ⇄ cC+dD The square brackets show
Kc= [C] ͨ [D] ͩ [products] . concentration in moldm¯³
[reactants]
[A] ͣ [B]ᵇ The indices are the numbers in front of
Equilibrium constant each substance in the chemical equation
C shows that its in terms of
concentration
If Kc has a value of 1 then the
Examples position of equilibrium is
H₂ (g) +I₂(g) ⇄ 2HI(g) halfway between the reactants
and products
The equilibrium constants are If Kc>1 then the reaction is
H₂ 0.140 moldm¯³ product favoured so there are
HI 0.320 moldm¯³ more products at equilibrium
I₂ 0.0400 moldm¯³ If Kc<1 then the reaction is
Kc= [HI(g)]² Kc= [0.320(g)]² .=18.3 reactant favoured so there are
[H₂(g)][I₂(g)] [0.14(g)][0.04(g)] more reactants at equilibrium
, Example from experimental results
N₂(g) + 3H₂(g) 2NH₃(g)
2.5mol of N₂ and 6.2 mol of H₂ were placed in a 5dm³ container & were left
to reach equilibrium. 2 mol of NH₃ was measured at equilibrium
Calculate the Kc of this reaction
Step 1: work out the equilibrium amounts of all equilibrium species using
MICE (Eq)
ratio 1 3 2 if this was a
Molecules N₂ H₂ NH₃ heterozygous equilibrium
only (g) & (aq)would be
Initial moles 2.5 6.2 0 included in working out
Change in moles -1.0 -3.0 +2.0
If value of v is unknown
Equilibrium moles 1.5 3.2 2.0
just use ‘v’ as this will
Equilibrium concentration 1.5/5=0.3 3.2/5=0.64 20/5=0.4 cancel out in the Kc
(c=n/v) expression
Step 2: write expression for Kc
Kc= [NH₃]² units= [moldm¯³]² .
[N₂][H₂]³ [moldm¯³] [moldm¯³]³
Kc= 0.4² .
0.3 x 0.64³ units=mol⁻²dm⁶
= 2.03
2.03>1 so product favoured
