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Let the equation of a reaction be as follows.

**A + B = Products**

According to law of mass action:

**Reaction Rate ∞ [A] x [B]**

**Reaction Rate = k [A] x [B]**

Here k is a **constant** called the **rate constant** of this reaction.

If [A] = 1 and [B] = 1 then the velocity of reaction = k

Hence, *the rate constant of a reaction is equal to the velocity of
the reaction when the reactants are unit molecular concentrations.
Therefore, the Rate constant is also called the specific reaction rate.*

The value of the rate constant depends on the nature and heat of the reaction. At constant temperature, the value of the rate constant of a reaction is constant and fixed. The value of the rate constant does not depend on the concentrations or pressures of the reactants.

** Effect of heat on Rate constant:** The effect of heat on rate constant is represented by the

Here k is the rate constant, A is the frequency factor, E_{a} is the energy of activation, R is the universal gas constant and T is the ultimate temperature. A and E_{a} are characteristics of the reaction. The values of A and E_{a} are always the same for any one reaction. The values of A and E_{a} are different for different reactions.

If T_{1} and T_{2} are rate constant k_{1} and k_{2} of a reaction on heat then according to Arrhenius equation:

It is clear that the rate constant of each reaction also increases with increasing temperature. Accordingly, the velocity of the reaction also increases. It has been found that at 10°C temperatures the velocity of most reactions doubles.

** Effect of catalyst on rate constants:** In the presence of positive catalyst, the value of energy of activation (E

Equilibrium constant

Let the equation of a reversible reaction be as follows.

**A + B ⇋ C + D**

According to law of mass action:

**Reaction Rate ∞ [A] x [B]**

**Reaction Rate = k _{1} [A] x [B]**

Where k_{1} is the rate constant of the forward reaction.

Like this

**Reaction Rate ∞ [C] x [D]**

**Reaction Rate = k2 [C] x [D]**

Where k_{2} is the rate constant of the backword reaction.

Once the chemical equilibrium has taken place –

Forward Reaction Rate = backward reaction Rate

**k _{1} [A] x [B] = k2 [C] x [D]**

**k _{1 }/ k_{2} = [C] x [D] / [A] x [B]**

**K _{c} = [C] x [D] / [A] x [B]**

K_{c} is called **equilibrium constant**. **The
equilibrium constant of a reversible reaction is equal to the ratio of
Rate constants of the forward reaction and the backward reaction**.

Molar concentrations of the products in the expression of K_{c} are written in the numerator and molar concentrations of the reactants in the denominator. The subscript c of K_{c} shows that molar concentrations are used to the reactants and products in the expression of K_{c}.

The expression of equilibrium constant represents the concentration
at the equilibrium of the reactants and products. Any time the product
of concentrates of products is divided by the product of the
concentrations of the reactants, the quotient is called **reaction quotient** and it is represented by Q.

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For the following general reaction –

**aA + bB ⇋ cC + dD**

If Q < Kc then the reaction takes place in the forward direction.

If Q > Kc then the reaction takes place in the backward direction.

If Q = Kc then the reaction occurs in equilibrium.

If the rate constant of the forward reaction of a reversible reaction is K_{1}, the rate constant of the backward reaction K_{2} and the equilibrium constant K_{c}, then-

**K _{c} = K_{1} / K_{2}**

Like the rate constant, the value of equilibrium constant also
depends on the nature and heat of the reaction. At constant temperature,
the value of the **equilibrium constant** of a reversible reaction
is constant and fixed. Like the rate constant, the value of equilibrium
constant also does not depend on the concentrations or pressures of the
reactants and products.

The presence of a catalyst changes the value of the rate constant. The presence of a catalyst has the same effect on forward and backward reactions and does not change the value of equilibrium constants.

For the following reversible reactions –

**A + B ⇋ C + D**

Equilibrium constant of forward reaction **(K _{c}) = [C][D]/[A][B]**

Equilibrium constant of opposite reaction **(K _{c}’) = [A][B]/[C][D]**

Therefore **K _{c} = 1/K_{c}^{’}**

Hence, the equilibrium constants of the forward and backward reactions of a reversible reaction are inverse to each other.

Equilibrium constant of forward reaction **(K _{c}) = [C][D]/[A][B]**

Equilibrium constant of opposite reaction **(K _{c}’) = [A][B]/[C][D]**

Therefore **K _{c} = 1/K_{c}^{’}**

Hence, the equilibrium constants of the forward and backward reactions of a reversible reaction are inverse to each other.

The equilibrium constants of chemical reactions depend on their equations. If the equation of a chemical reaction is written in a different form, the value of its equilibrium constant will change.

** Example:** The equation of thermal dissociation of nitrogen dioxide can be written as:

**2NO _{2} ⇋ N_{2} + 2O_{2}**

For the above reaction –

Equilibrium constant (**K _{c}) = [N_{2}][O_{2}]^{2} / [NO_{2}]^{2}**

The equation of thermal dissociation of hydrogen dioxide can also be written as-

**NO _{2} ⇋ ½ N_{2} + O_{2}**

For the above reaction-

Equilibrium constant **(K _{c}’) = [N_{2}]^{1/2}[O_{2}]/[NO_{2}]**

According to equations 1 and 2 –

**K _{c} = (K_{c}’)^{2}**

It is clear that according to both the above equations different values of equilibrium constants are obtained, that is, the value of equilibrium constant depends on the equation of reaction.

Therefore, to write the value of equilibrium constant one should use the standard equation of reaction or the given equation, the equation should not be changed by multiplying or dividing the number of molecules represented in the equation.

According to the ideal gas equation –

**PV = nRT**

**P = n/V x RT**

**P = CRT**

Where **C = n / V** = number of moles / volume in liters = molar concentration

The partial pressure of a gas is proportional to its **molar concentration** by constant heating according to the above equation.

For this reason, partial expressions of gases can be used instead of
molar concentrations in the expression of equilibrium constants of
gaseous reactions. In this case, the equilibrium constant is represented
by **K _{p}**.

Let the equation of a reversible gaseous reaction be as follows –

**A(g) + B(g) ⇋ C(g) + D(g)**

Using the law of mass action on this reaction, the values for the reaction constants K_{c} and K_{p} will be as follows –

**K _{c} = [C] x [D] / [A] x [B]**

**K _{p} = P_{C} x P_{D} / P_{A} x P_{B}**

P_{A}, P_{B}, P_{C,} and P_{D} are partial pressures of A, B, C, and D at equilibrium, respectively.

Let the equation of a reversible gaseous reaction be

as follows:

**aA(g) + bB(g) ⇋ cC(g) + dD(g)**

In the above reaction, A and B are reactants and C and D are products. a, b, c and d are the numbers of their molecules.

All reactants and products are in gaseous state. Using the law of mass action on this reaction, the values of K_{c} and K_{p} will be as follows –

**K _{c} = [C]^{c} x [D]^{d} / [A]^{a} x [B]^{b}**

**K _{p} = P_{C}^{c} x P_{D}^{d} / P_{A}^{a} x P_{B}^{b}**

According to the ideal gas equation-

**PV = nRT**

**P = n/V x RT**

P = CRT where C is **molarity **

P_{A} = C_{A}RT = [A] x RT

P_{B} = C_{B}RT = [B] x RT

P_{C} = C_{C}RT = [C] x RT

P_{C} = C_{D}RT = [D] x RT

Put the value of P_{A}, P_{B}, P_{C} and P_{D }in the equation

**K _{p} = [C]^{c} (RT)^{c} [D]^{d} (RT)^{d} / [A]^{a} (RT)^{a} [B]^{b} (RT)^{b}**

**K _{p} = [C]^{c} x [D]^{d} / [A]^{a} x [B]^{b} x (RT)^{c+d-a-b}**

**K _{p} = K_{c} x (RT)^{Δ}^{n}**

Where **Δ**n there is a *difference of the number of molecules of the products and the reactants*.

The expression of equilibrium constant K_{c} contains the terms of molar concentrations. Molar concentrations are expressed in mole per liter.

Hence the **unit of K _{c} is (mol/liter)**

*Where Δn = number of molecules of products – number of molecules of reactants*

Similarly, the expression of equilibrium constant K_{p} includes pressure terms. The pressure is expressed in the atmosphere, so the unit of **K _{p} is (atmosphere)**

It is clear that the unit of K_{c} and K_{p} depends on the type of reaction.

If **Δ**n = 0 i.e. the number of molecules of the products is equal to the number of molecules of the reactants then K_{p} = K_{c} and K_{p} and K_{c} have no units.

The value of R is 0.0821 liter-atmosphere mole^{-1} K^{-1}. The value of T is usually around 300 K so the value of RT is greater than 1.

Therefore, if **Δ**n is positive, that is, the number of molecules of the products is greater than the number of molecules of the reactants, then K_{p} > K_{c}. If **Δ**n is negative, that is, the number of molecules of the reactants is greater than the number of molecules of the products, then K_{p} < K_{c}.

**Following are examples of reactions in which the number of molecules changes.**

**1) N _{2}(g) + 3H_{2}(g) ⇋ 2NH_{3}(g)**

K_{c} = [NH_{3}]^{2} / [N_{2}][H_{2}]^{3}

Unit of K_{c }= (mol per liter)^{-2}

= liter^{-2} mol^{-2}

K_{p} = PNH_{3}^{2} / PN_{2} x PH_{2}^{3}

Unit of K_{p} = (atmosphere)^{-2}

**2) 2NO _{2}(g) ⇋ N_{2}(g) + 2O_{2}(g)**

Kc = [N_{2}][O_{2}]^{2} / [NO_{2}]^{3}

Unit of K_{c }= (mol per liter)^{1}

= mol per liter

K_{p} = PN_{2} x PO_{2}^{2} / PNO_{2}^{2}

Unit of K_{p} = (atmosphere)^{1}

Equilibrium constant K_{c} = rate constant of forward reaction(k_{1}) / rate constant of opposite reaction(k_{2})

**The value of rate constant increases with increasing temperature.**

Therefore, the values of k_{1} and k_{2} increase with increasing temperature. If the value of k_{1} increases in greater proportion then the value of K_{c} will increase. If the value of k_{2} increases in greater proportion then the value of K_{c} will decrease.

Therefore, the value of rate constant always increases with increasing temperature but the value of equilibrium constant can also increase and decrease.

**Van’t Hoff **studied the effect of heat on **equilibrium constants** based on the principles of **thermodynamics **and established the following relations –

Where T1 and T2 are equilibrium constants (Kp)1 and (Kp)2 at absolute temperature, R is the universal gas constant and ΔH is reaction heat of forward-reaction in the constant pressure.

Where T1 and T2 are equilibrium constants (Kc)1 and (Kc)2 at absolute temperature, R is the universal gas constant and ΔE° is the reaction heat of the forward reaction at constant volume. For liquid systems, one can also write ΔH° in place of ΔE° in the above equation.

Both the above relations can also be obtained with the help of arrhenius equation. In fact arrhenius equation is an empirical equation.

Heat is absorbed in endothermic reactions and the value of ΔH° is positive for these. It is clear that the value of equilibrium constant increases with increasing temperature of endothermic reactions.

Exothermic reactions emit heat and for these the value of ΔH° is negative. It is clear that the value of equilibrium constant decreases with increasing temperature of exothermic reactions.

** Calculation of equilibrium concentrations:** If the
initial concentrations of the reactants are known, the remaining
quantities of the reactants and the quantities of the products can be
calculated with the help of equilibrium constants.

** Measuring the extent of the reaction:** With the help of
the value of equilibrium constant, it can be told to what extent the
authorities have converted into products before the equilibrium was
established.

**example:**

The value of the equilibrium constant of the reaction at 1000 K is 2.2×10^{22}.

**2CO(g) + O _{2}(g) ⇋ 2CO_{2}(g)**

It is clear that the concentration of CO_{2} at equilibrium is much higher than the concentrations of CO and O_{2},
that is, the reactants have been converted into almost complete
product. Therefore, the higher the value of K, the greater the tendency
for the reaction to be in the forward direction.

__ Stability of products:__ With the help of the value of
equilibrium constant, the relative stability of the reactants and
products is obtained. If the value of K is high then the products are
more permanent and if the value of K is less then the products are less
permanent.

**Example**:

**N2 + O _{2} ⇋ 2NO K = 5 x 10^{-31}**

**N2 + 2O _{2} ⇋ 2NO_{2} K = 2 x 10^{-31}**

Comparing the equilibrium constants of these

reactions, it is found that NO is more stable than NO_{2}.