# EQUILIBRIA CONCENTRATIONS

EQUILIBRIA REACTIONS

Reversible reactions occur in which reactants change to products while products convert back to form reactants.

Reversible reactions can reach equilibria state or point. This is the state at which constant amounts of reactants and products exist. Amounts of product formed from reactants is equal to amount of reactants formed back from products.

A ^{______}> C or A + B ^{_______}> C non-equilibrium reaction, goes completely to products

A <= > C or A + B < = > C equilibrium reaction, reaches equilibrium

At equilibrium point, concentration or moles of A and B that react to form product C are equal that is;

[A]eq = [B]eq = [C]eq

[A]eq = Concentration of A that reacts with B = moles of A lost/Vrxn

[B]eq= Concentration of B that reacts with A = moles of B lost/Vrxn

[C]eq= Concentration of product C formed = moles of C formed/Vrxn

Vrxn = reaction volume in liters

Equilibrium constant = [Amount of product] / [amount of reactants] = Constant = K at temperature T

**K = [C]rem / ([A]rem x [B]rem)**

[A]eq = [B]eq implies amount of A and B at equilibrium that react to form product C are equal at equilibrium.

[A]rem = [A]i – [A]eq

where [A]i = initial amount concentration of A at beginning of reaction

[A]rem = Amount concentration of A left that remains unreacted at equilibrium

[A]eq = [A]i – [A]rem = initial moles of A/Vrxn – moles of A remaining at equilibrium/Vrxn

Vrxn = total volume of reaction in liters

For an equilibrium reaction A + 2B < = > C

This means twice the moles of B reacted with A to form C.

Moles of A lost = Aeq= C moles

Moles of B lost = Beq= 2C moles

Concentration of A remaining = [Arem] = [(Ai – Aeq)/V], where V is reaction volume in Liters.

Concentration of B remaining = Brem= [(Bi – Beq)/V]

[A]rem = [B]^{2}rem =[C]

**EQUILIBRIUM CONSTANT = K**= [C]/ ([A]rem x [B]^{2}rem) ^{ }

EXAMPLE

In a reaction of A < = > B

Calculate the equilibrium constant of reaction if 2 moles of A converted to 0.5 moles of B at 100^{o}C in a reaction volume of 100mL.

Moles of A lost = moles of B formed = 0.5 mol

Moles of A remaining = initial moles of A – Moles of A lost

= 2 – 0.5 = 1.5 mol

K = Concentration product/ Concentration reactant

K= (0.5mol/100 x 10-3L) / (1.5mol /100 x 10-3L) =0.5mol/1.5mol = 0.33

How many moles of B will be produced from 5.5 mol of A in a reaction volume of 250mL at 100^{o}C.

0.33 = [B] / [5.5-B]

0.33[5.5 – B] = B

1.815-0.33B=B

1.815 = 1.33B

B= 1.815/1.33 = 1.36 mol

If 5.0mol of A was mixed with 5.0mol of B calculate the concentration of B when reaction reaches 100^{o}C.

The reaction reach equilibrium at 100^{o}C, K= 0.33

A rem = A initial – B formed = 5.0 – Bf

B rem = B initial + B formed=5 + Bf

0.33 = (5.0+ Bf)/ (5-Bf)

0.33 x (5 – Bf) = 5 + Bf

1.65-0.33Bf = 5 + Bf

1.65-5=1.33Bf

-3.35 = 1.33Bf

Bf = -3.35/1.33 = -2.5mol means 2.5mol B lost to form A at equilibrium

A rem = 5- – 2.5 = 7.5mol

B rem = 5 + – 2.5 = 2.5 mol , Concentration = 2.5 mol / 250mL = 10M

In the reaction A + B < = > C,

2.5 mole of A was mixed with 4 moles of B and 0.59 mole of C product was formed at 25 ^{o}C in a reaction vessel of volume 250mL. What is the value of the equilibrium constant?

Moles of A lost = moles of C formed = moles of B lost

Moles of A remaining = 2.5 – 0.59 = 1.91 mol, Concentration A = 1.91mol/[250 x10^{-3}L] =7.64M

Moles of B remaining = 4 – 0.59 = 3.41 mol, Concentration B =3.41mol /[ 250 x 10^{-3}L]= 13.64M

Moles of C formed = 0.59 mol, Concentration C = 0.59 mol / [250 x 10^{-3}L] = 2.36 M

K = [product]/[reactants] = [Concentration C]/ [(Concentration A x Concentration B)]

= [(2.35M)] / [7.64M x 13.64 M] = 0.0226 M^{-1}

If the above reaction was carried out with 12 moles of A and 5 moles of B? How moles of product C will be formed in reaction volume of 1.0 L.

A rem= 12.0 – C, Concentration A rem = (12- C)/1.0L

B rem = 5- C, Concentration B rem = (5 – C) / 1.0 L

K = [C]/ [(12 – C) x (5 – C)]

0.0226 M^{-1} = C / (60 -12C -5C + C^{2})

0.0226 M (60 – 17C + C^{2}) = C

1.356 – 0.3842 C + 0.0226 C^{2} = C

1.356-0.3842C +0.0226C^{2} – C = 0

1.356 – 1.3842 C + 0.0226 C^{2} = 0

**Quadratic equation ax ^{2} + bx + c = 0,**

FIND X= C

X = (-b +/- (b^{2} – 4ac)^{0.5})/2a

X = – (-1.3842) +/- [(-1.3842^{2}) – 4 x 0.0226 x 1.356]^{0.5} / (2 x 0.0226)

X = 1.3842 + (1.916-0.1226)^{0.5} / 0.0452 or 1.3842 – (1.916-0.1226)^{0.5} / 0.0452

X = (1.3842+1.339)/0.0452 or (1.3842 – 1.339) / 0.0452

X= 60.25M or 1.0M (more reasonable)

1.0M in 1.0L contains 1.0 mol of C

**REACTION QUOTIENT**

Q = Concentration Products/ Concentration Reactants= Reaction quotient = Direction of reaction = Forward right products or backwards to left reactants

Q > K = Back to Reactants

Q < K = Forward to Products

Q = K = Equilibrium point Q = K [amount of reactants and products becomes constant]

EXAMPLE

In a reaction, A + B < = > C, there is 2 mole of A, 3 Mole of C and 0.5 mole of B mixed in reaction vessel of volume 2.4 L. Predict whether the reaction will go to the left backwards or forward right to form products at that temperature.

Q = [C]/([A][B]) = (3/2.4)/([2/2.4] x[0.5/2.4]) = 1.25/(0.833 x 0.20833) = 7.2

Q is larger than K {0.0226}, therefore the reaction will NOT go forward to form products. It will go backward to the left to form reactants till equilibrium is attained.

Calculate all the values of the above reaction situation for a reaction of A + 4B ^{________}> 2C

Moles of A lost = 2 moles of C formed = 4 moles of B lost

Moles of A remaining = 2.5 – (1/2) x 0.59 = 2.205mol, Concentration A = 2.205mol/ [250 x10^{-3}L] =8.82M

Moles of B remaining = 4 – (4/2)0.59 = 2.82 mol, Concentration B =2.82 mol / [ 250 x 10^{-3}L] = 11.28M

Moles of C formed = 0.59 mol, Concentration C = 0.59 mol / [250 x 10^{-3}L] = 2.36 M

K = [product]/[reactants] = [Concentration C]^{2}/ ([Concentration A] x [Concentration B]^{4})

= [(2.35M)]^{2} / ([8.82M] x [11.28 M]^{4}) = 0.004921 M^{-3}

If the above reaction was carried out with 12 moles of A and 5 moles of B? How moles of product C will be formed in reaction volume of 1.0 L.

A rem= 12.0 – ½ C, Concentration A rem = (12- 0.5C)/1.0L

B rem = 5- ^{4}/_{2 }C, Concentration B rem = (5 – 2C) / 1.0 L

K = [C]^{2}/ [(12 – 0.5 C) x (5 – 2C)^{4}]= 0.004921 M^{-3}

K = [C]^{2}/ [(12 – 0.5 C) x (5 – 2C)(5-2C)(5-2C)(5-2C)]= 0.004921

K = [C]^{2}/ [(12 – 0.5 C) x (25 – 10C-10C+4C^{2})^{2}]= 0.004921

K = [C]^{2}/ [(12 – 0.5 C) x ( 25– 20C+4C^{2})( 25– 20C+4C^{2})]= 0.004921

[C]^{2}/ [(12 – 0.5 C) x ( 625– 500C+100C^{2}-500C+ 400C^{2}– 80C^{2}+100C^{2}+80C^{2}+16C^{4})]= 0.004921

Solve for C (long process.)

However, Let’s assume that C is small, therefore C^{2}, C^{3} or C^{4} will be very small and therefore cancel out.

[C]^{2}/ [(12 – 0.5 C) x ( 625– 1000C)] =0.004921

[C]^{2}/ [(7500 – 12000C- 312.5C+ 500C^{2 }]= 0.004921

[C]^{2} = 0.004921[ 7500 – 12000C- 312.5C+ 500C^{2 }]

[C]^{2} = 36.9 – 59.052C- 1.5378C+ 2.46C^{2 }] = 36.9 – 60.60C + 2.46C^{2}

0=36.9 – 60.60C + 1.46C^{2}

You can solve for C again with quadratic equation or cancel out 1.46C^{2}

0=36.9 – 60.60C

then 36.9=60.60C

C= 36.9/60.60 = 0.61 M or 0.61 moles in 1.0 L

If 2 mole of A, 3 Mole of C and 0.5 mole of B mixed in reaction vessel of volume 2.4 L. Predict whether the reaction will go to the left backwards or forward right to form products at that temperature.

Solve for Q

Q = [C]^{2} / ([A] x [B]^{4}) = (3/2.4)^{2}/([2/2.4] x [0.5/2.4]^{4}) = 995.328

Q >>> K implies reaction may not form any product. C will convert to A and B till equilibrium is attained.