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**Equilibrium constant**

aA + bB cC + dD

At equilibrium, R_{f} = R_{b}

k_{f}[A]^{a} [B]^{b} = k_{r}[C]^{c} [D]^{d} →

K_{c} → equilibrium constant in terms of concentration

(K_{p} = eq. constant in terms of partial pressure)

for the reaction, N_{2}(g) + 3 H_{2}(g) 2NH_{3}(g)

,

**Law of Chemical Equilibrium**

At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the equilibrium Law or Law of Chemical Equilibrium

**Units of Equilibrium Constant**

The value of equilibrium constant K_{C} can be calculated by substituting the concentration terms in mol/L and for K_{p} partial pressure is substituted in Pa, kPa, bar or atm. This results in units of equilibrium constant based on molarity or pressure. unless the exponents of both the numerator and denominator are same.

For the reactions,

H_{2}(g) + I_{2}(g) 2HI, K_{C} and K_{P} have no unit.

N_{2}O_{4}(g) 2NO_{2} (g), K_{C} has unit mol/L and K_{P} has unit bar or atm

Equilibrium constants can also be expressed as dimensionless quantities if the standard state of reactants and products are specified. For a pure gas, the standard state is 1 bar. Therefore a pressure of 4 bar in standard state can be expressed as 4 bar/1 bar = 4, which is a dimensionless number. Standard state (C_{0}) for a solute is 1 molar solution and all concentrations can be measured with respect to it. The numberical value of equilibrium constant depends on the standard state chosen. Thus in this system both K_{p} and K_{C} are dimensionless quantities and represented as K_{p}° & K_{C}° respectively

**Relation between K _{p} and K_{c}**

For the reaction aA + bB cC + dD

**K _{p}=K_{C}(RT)^{Dng}**

where Δn_{g} = (c +d) - (a +b) = no. of moles of gaseous products - no. of moles of gaseous Reactants

if Δn = 0 K_{p} = K_{c}

**Significance of Equilibrium Constant:**

**(a) Using K _{eq} to Predict Relative Concentrations**

The size of the equilibrium constant can give us information about the relative amounts of reactants and products present at equilibrium.

• **When K << 1**

The reaction lies to the left (mostly reactants)

• **When K >> 1**

The reaction lies to the right (mostly products)

• **When K = 1**

The reaction lies in the middle (mix of reactants and products)

**(b) Calculating Equilibrium concentrations**

**Ex. Phosgene is a poisonous gas that dissociates at high temperature into two other poisonous gases, carbon monoxide and chlorine. The equilibrium constant K _{p}= 0.0041 atm at 600K. Find the equilibrium composition of the system after 0.124 atm of COCl_{2} initially is allowed to reach equilibrium at this temperature.**

**Sol.**

| COCl | ||

initial pressures | 0.124 | 0 | 0 |

change | -x | +x | +x |

equilibrium pressures | 0.124-x | x | x |

**Substitution of the equilibrium pressures into the equilibrium expression gives**

**This expression can be rearranged into standard polynomial form**x^{2} 0041x-0.00054=0 and solved by the quadratic formula, but we will simply obtain an approximate solution by iteration.

Because the equilibrium constant is small, we know that x will be rather small compared to 0.124, so the above relation can be approximated by

which gives x=0.025. **To see how good this is, substitute this value of ***x *into the denominator of the original equation and solve again:

This time, solving for x gives 0.0204. Iterating once more, we get

and** ***x *= 0.0206 which is sufficiently close to the previous to be considered the final result. The final partial pressures are then 0.104 atm for COCl_{2}, and 0.0206 atm each for CO and Cl_{2}.

**Note:** using the quadratic formula to find the exact solution yields the two roots -0.0247 (which we ignore) and 0.0206, which show that our approximation is quite good. *Note*: using the quadraticformula to find the exact solution yields the two roots -0.0247 (which we ignore) and 0.0206, which show that our approximation is quite good.

**Reaction Quotient (Q):**

At each point in a reaction, we can write a ratio of concentration terms having the same forms as the equilibrium constant expression. The ratio is called the reaction quotient denoted by symbol Q. It helps in predicting the direction of a reaction.

The expresson Q= at any time during rection is called reaction quotient

(i) if Q > K_{C }reaction will proceed in a bakward direction until equilibrium in reached.

(ii) if Q < K_{C }reaction will proceed in forward direction until equilibrium is established.

(iii) if Q = K_{C} reaction is at equilibrium

**Ex. For the reaction NOBr (g) NO(g) + 1/2 Br _{2}(g) , K_{p}=0.15 atm at 90°C. If NOBr, NO and Br_{2} are mixed at this temperature having partial pressures 0.5 atm,0.4 atm & 2.0 respectively, will Br_{2} be consumed or formed?**

**Sol.**

K_{P}=0.15

Hence, reaction will shift in backward direction, Therefor Br_{2} will be consumed

**Degree of Dissociation & Vapour Density**

*Ex. The Vapour Density of mixture of PCl _{5}, PCl_{3} and Cl_{2} is 92. Find the degree of dissociation of PCl_{5}.*

**Sol.**

** **

1 0 0

1 - x x x

= 104.25 ,

, α** =0.13**

**Characteristics of equilibrium constant**

**1**.

After reversing the reaction

**After reversing the reaction the equilibrium constant get reversed.**

**2**.

and

A + B + E + F C + D + G K_{3}

→ K_{3} = K_{1} × K_{2}

**when the two reaction are added there equilibrium constant get multiplied.**

**3**.

After multiplying by n

nA + nB nC + nD K_{2}

**When the reaction is multiplied by any number then eq. constant gets the same number in its power.**

**Ex. Given the following equilibrium constants:**

**(1) CaCO _{3}(s) → Ca^{2+ }(aq) + CO_{3}^{2-}(aq) K_{1}=10^{-8.4}**

**(2) HCO _{3}^{-}(aq) → H^{6}^{ }(aq) + CO_{3}^{2-}(aq) K_{2}=10^{-10.3}**

**Calculate the value of K for the reaction CaCO _{3}(s) + H^{ }(aq)=Ca^{2 }(aq) + HCO_{3}^{-}(aq)**

**Sol.** The net reaction is the sum of reaction 1 and the reverse of reaction 2:

CaCO_{3}(s) → Ca^{2 }(aq) +CO_{3}^{2-}(aq)** **K_{1}=10^{-8.4}

H^{+}^{ }(aq) + CO_{3}^{2-}(aq)→ HCO_{3}^{-}(aq) K_{-2}=10^{-(-10.3)}

^{-------------------------------------------------}

CaCO_{3}(s) + H^{+}^{ }(aq) → Ca^{2 }(aq) + HCO_{3}^{-}(aq), K=K_{1}/K_{2}=10^{(-8.4 +10.3) }=10^{+}^{1.9}

*Comment: *This net reaction describes the dissolution of limestone by acid; it is responsible for the eroding

effect of acid rain on buildings and statues. This is an example of a reaction that has practically no tendency to take place by itself (the dissolution of calcium carbonate) begin "driven" by a second reaction having a large equilibrium constant. From the standpoint of the LeChâtelier principle, the first reaction is "pulled to the right" by the removal of carbonate by the hydrogen ion. "Coupled" reactions of this type are widely encountered in all areas of chemistry, and especially in biochemistry, in which a dozen or so reactions may be linked in this way.

**Ex.** **The synthesis of HBr from hydrogen and liquid bromine has an equilibrium constant Kp = 4.5×10 ^{15} at 25°C. Given that the vapor pressure of liquid bromine is 0.28 atm, find Kp for the homogeneous gas-phase reaction at the same temperature.**

**Sol**: The net reaction we seek is the sum of the heterogeneous synthesis and the reverse of the vaporization of liquid bromine:

Here are some of the possibilities for the reaction involving the equilibrium between gaseous water and its elements:

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