Opposing Reactions

Opposing reactions

They are reversible reactions in which a product is formed, reacts and give the reactant(s) again. Such reactions are represented by a double headed arrow (⇆).

At start rate of forward reaction (Rf) is high and rate of backward reaction (Rb) is slow. With time, Rf decreases and Rb increases. At some point, the rate of forward and backward reaction becomes equal but, at that point the concentration of reactants and products are not equal. For example, for a reaction b/w H₂ and I₂. consider following reaction concentration of products and reactants and time.

H₂ + I₂ ⇆ 2HI

Time (minute)	Reactants	Products
0	10	0
5	9	2
10	8	4
15	7	6
20	6	8
25	6	8

Graphically,

Similarly, for the reversible reaction consider following time and concentrations.

2HI → H₂ + I₂

Time (minute)	Reactants	Products
0	10	0
5	8	1
10	6	2
15	4	3
20	2	4
25	2	4

Graphically,

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1^st order opposed by 1^st order case I

These are the reactions in which both the rate of forward and backward reaction is of first order. Case I involves reactions having no initial products.

R ⇄ P

R P

At time t=0 a 0

At time t a-x x

At equilibrium a-xe xe

Where xe is concentration of product at equilibrium. Now we know that

Rate of forward reaction Rf ∝ (a-x)

Rf = kf (a-x)

Rate of backward reaction Rb ∝ (x)

Rb = kb (x)

Net rate of reactions will be dx/dt = Rf - Rb

dx/dt = kf (a-x) - kb (x) eq. 13.1

above is differential rate expression for reversible reaction. the values of both constants kf an kb can be found as follows. As we know that at equilibrium rate becomes constant so, net rate will be zero i.e. dx/dt = 0. So, eq. 13.1 becomes

0 = kf (a-x) - kb (x)

Or kf (a-x) = kb (x) eq. 13.2

At equilibrium eq.13.2 becomes

kf (a-xe) = kb (xe) eq. 13.3

where xe is the concentration of product at equilibrium. From eq.13.3

eq. 13.4

Putting the above value of kb in eq. 13.1 we got

by taking LCM

eq. 13.5

here is constant so, eq. 13.5 becomes

integrating the above equation we got

eq. 13.6

to find the value of z considering x=0 and t=0 and putting in eq. 13.7 we got

z = -ln (xe)

putting the value of z in eq. 13.6 we got

eq. 13.7

above equation is a straight line equation of opposing reaction 1st order opposed by 1st order having no initial concentration of products. Graphically,