Chemical Equilibrium

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1
Kequilibrium =?
[C]^c*[D]^d/([A]^a*[B]^b)
2
relation b/w kp and kc
Kp=kc(RT)^∆n
3
equilibrium constant in terms of mole fraction
Kx=Xx^x*Xy^y/XA^a*XB^b ,Kp=Kx(P)^∆n
4
CHARACTERISTICS OF EQUILIBRIUM CONSTANT
· Value of K does not depend upon initial concentration of reactants and products.
· Value of K does not depend upon the direction from which equilibrium is attained.
· Value of K depend only on temperature.
· If the equilibrium constant for the reaction aA + bB ightleftharpoons cC + dD is K, then the equilibrium constant for the reaction cC + dD ightleftharpoons aA + bB will be 1/K
· If the equilibrium constant for the reaction aA + bB ightleftharpoons cC + dD is K, then the equilibrium constant for the reaction naA + nbB ightleftharpoons ncC + ndD will be (K)ⁿ
· During the addition of two reaction having equilibrium constants K₁ and K₂, then the net equilibrium constant K = K₁ × K₂
· During the subtraction of a reaction having equilibrium constants K₂ from a reaction having equilibrium constant K₁, then the net equilibrium constant K K1/K2,If Kc > 10³, products predominate over reactants, i.e., if Kc is very large, the reaction proceeds nearly to completion.
· If Kc < 10⁻³, reactants predominate over products, i.e., if Kc is very small, the reaction proceeds only to a very small extent.
· If Kc is in the range of 10⁻³ to 10³, appreciable concentrations of both reactants and products are present.
5
dissociation of a weak acid
Ka=C*(aplha)^2/(1-aplha) since alpha<<1 therefore Ka=C*(alpha)^2 or alpha=√(Ka/C)
6
relative strength of acidsases
for acids equal concentration HA1/HA2 =√(Ka1/Ka2) , for acids having different concentrations HA1/HA2 = √(Ka1*C1/Ka2*C2)
7
dissociation of poly protic acid
Ka=Ka1*Ka2,Ka1>Ka2, consider a triprotic acid Ka=ka1×Ka2×Ka3
8
Kw=?
Kw=[H3O+][OH-],pKw=-log Kw, for pure water at 298K Kw= 10^-14 pKw=14 at T=273K pKw=15 ,at 333K pKw=13
9
pH of weak acids and bases
[H3O+]=c*alpha=Ka/aplha=√(ka×c),
10
pH of a buffer solution
pH of an acidic buffer pH=pKa+log([salt]/[acid]) pH=pKa+log([conjugated base/[acid]). pH of a basic buffer pOH=pKb+log([salt]/[base]) pH=pKa+log([conjugated acid/[base]).
11
buffer capacity
buffer capacity=no.ofmoles of acid or base added per litre of buffer/change in pH
12
hydrolysis of salts
salt of WA+SB(Anionic hydrolysis)((Kh=Kw/Ka ,h=√(Kh/C) ,pH=1/2(pKw+pKa+log C),nature of solution is basic) salt of WB+SA(cationic hydrolysis) Kh=kW/kb ,h=√(kh/C) ,pH=1/2(pKw-pkb-logC) nature of solution is acidic), salt of WA+WB(both Anionic and cationic hydrolysis) (kh =kw/ka*kb , h=√Kh ,pH=1/2(pKw+pKa-pKb)acidic r basic or neutral) salt of SA+SB(No hydrolysis) pH=7 nature of solution is neutral)
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