REVIEW

CENTRAL CHEMISTRY:

Central Chemistry(or Infinitesimal Chemistry) is a spacial branch of Chemistry that applies the Central Point Law, and Point Values Interval Law. 

POINT RELATIVE ATOMIC MASSES

The mass spectrum of element will show up the presence of isotopes: from the relative height atomic mass of the elements. For example, any naturally occurring sample of the elements ^AE and ^BE in the proportions P₁:P₂, there is corresponds prediction of P₁ and P₂ at a given equation A(P₁/100)+B(P₂/100)=A_r, where A_r is atomic mass;

P₁=A_r100/2A_r

P₂=A_r/2B

EXAMPLE:

Any naturally occurring sample of chlorine contains ³⁵Cl and ³⁷Cl in proportions P₁:P₂. Predict the proportion P₁ and P₂ at relative atomic mass 35.50.

SOLUTION

A(P₁/100)+B(P₂/100)=A_r

P₁=35*1OO/2*35=50.714

P₂=35*1OO/2*37=47.973

PROOF!

35*(50.714/100)+37*(47.973/100)=35.50

17.7499+17.75001=35.50

35.49991=35.50

POINT MOLE REFRACTION AND VAPOR PRESSURE OF COMPONENTS

Usually decrease in the vapor pressure of a solvent depends on the number of solute particles in the solution, but on the nature of the solute in ideas solutions. Dilute solutions is related to number of solute particles and not merely on the nature of the solute. The ratio of the number of moles of final component in solution to four: all times one divided by the initial component in solution plus one divided by the final component in solution is called point fraction of a component I, xi can be expressed as;

X_i=n_i/4[1/n₁+1/n_i]

Also, for an ideal solution, where the vapor pressure of a given component in solution is directly proportional to the mole fraction of that component, the total equation of this relation is;

P_i=X_iP^o_i

P^o_i=P_s/2X_i

Where pi is the vapor pressure of component I in solution at a given temperature, X_i the mole fraction of component I and P0i the vapor pressure of the same temperature as the solution. From the above formulas, if P_A=X_AP_A and P_B=X_BP_B, the vapor pressure of the solution, P_s, is the same of the vapor presence of the components A and B in solution.

P_s =(P_A=P_B)+(P_A=P_B)=X_AP^o_A+X_BP^o_B

P^o_A=P_s/2X_A

P^o_B=P_s/2X_B

EXAMPLE:

What is the point fraction, and vapor pressure of pure components, heptanes (C₇H₁₆) and octane (C₈H₁₈ ) of a solution that contains 3.00mol of heptanes (C₇H₁₆) and 5.00 mol of octane(C₈H₁₈) at 40^oc, assuming ideal behavior? The vapor pressure of the two components is 53.

X_C7H16=n_C8H18[1/n_C7H16+1/n_C8H18]

X_C7H16=3.00/4[1/5.00+1/3.00]

X_C7H16=0.75[0.2+0.3333333333]

X_C7H16=0.4

X_C8H18=1-X_C7H16

X_C8H18=0.6

But;

X_C7H16P^o_C7H16+X_C8H18P^o_C8H18=P_solution

0.4P^o_C7H16+0.6P^o_C8H18=53

P^o_C7H16=53/2(0.4)=66.25

P^o_C8H18=53/2(0.6)=44.16666667

P_C7H16=0.4*66.25=26.5

P_C8H18=0.6 *44.166666667=26.5

PROOF

P_C7H16+P_C8H18=P_s

26.5+26.5=53

53=53

POINT STANDARD ENTROPY CHANGE OF REACTION FROM ABSOLUTE TEMPERATURE

Considered the reactants of chemical reaction that defined the initial state of a system and the products as the final state of the system. The converting  to products in their standard state depend on the reactants and the standard entropy change of the reaction equals the standard entropy of the final states (the products) minus the standard entropy of initial state(the reactants). Given the reaction equation:

Aa+bB→Cc+Dd

∆S^o_rxn=(CSoc+dSoD)-(aSoA+bSoB)

S^o_C=[∆S_rxn+(aSoA+bSoB)]/2c

S^o_D=[∆S_rxn+(aSoA+bSoB)]/2d

EXAMPLE:

Calculate the standard entropy change ∆S^o, for the reaction 2KCIO₃(s)+2KCI(s)→2KCI(s)+3O₂(g), using the absolute entropy value given as; S^o_KCIO3(s)=143mol^-1K^-1.

[Take ∆S_rxn =494K^-1]

SOLUTION

S^o_KCI=(494+143)/2(2)

S^o_KCI=159.25

S^o_O2=(494+143)/2(3)

S^o_O2=106.17

POINT CELL E.M.F.S THEORY

Cell e.m.f.s change when the concentration of the ions change;  a change in concentration of the solution in either the element A or the element B cell, might keep the A ion concentration fixed or vary at 1moldm^-3.The cell e.m.f.s equation are;

nfE_{A ion/A}-nfE^ө_{A ions/A}=RTln[A ion concentration (aq)]

nfE_{B ion/B}-nfE^ө_{B ions/B}=RTln[B ion concentration (aq)]

Where R is denoted as rate constant, F is denoted as faraday's constant , and n is number of moles. By illustration, if you set up a cell and change the concentration of the solution in either the zinc or copper half-cell, you might keep the zinc ion concentration fixed at 1moldm^-3 and used solution of copper(    II) sulphate with concentration varying  in modm^-3. E_cu2+/cu increase with increasing ln(cu²⁺ ) and potential changes as the concentration of cu²⁺ ion varies.

nfE_zn2+/zn-nfE^ө_zn2+/zn=RTln[zn²⁺(aq)]

nfE_cu2+/cu-nfE^ө_cu2+/cu=RTln[cu²⁺(aq)]

The zinc ion concentration is constant at 1 moldm^-3. The illustration above, with Central Point Law, has the cell equations.

E_cel=E_cu2+/cu - E^ө_zn2+/zn

E_cell=RT/2nf[ln(cu²⁺⁽aq)) + ln(zn²⁺(aq))]

RT=2nfE_cell/[ln(cu²⁺aq)+ln(zn²⁺aq)]

POINT POTENTIAL CURVE MODEL

The potential energy at a separation r, the term A/r¹² corresponds to repulsion (increase energy) and term –B/r⁶ tells about the attractions (lowering energy). That is to say, when molecules completely independently; but when they approach closely, the electron clouds repel one another. There is corresponds prediction of r¹² and r⁶ at a given equation V(r)=A/r¹² - B/r⁶, where A and B are constants.

r¹²=2A/V(r)

-r⁶=2B/V(r)

  

 POINT EQUILIBRIUM REACTION 

The equilibrium reaction law states that half the energy available for reaction is the energy of doing work and twice the heat change plus twice the energy of doing work, all divided by two is the energy available for reaction.

Twice the heat change is called point-energy available for reaction and twice the energy of doing work is another point-energy available for the reaction. 
∆S^ѳ=∆H^ѳ/2T=∆G^ѳ/T=[∆H^ѳ+∆H^ѳ]/2T=[T∆S^ѳ-RtlnK]/T=[T∆S^ѳ-∆G^ѳ]/T

EQUILIBRIUM CONSTANT

There are about four factors that influence equilibrium constant and these are:

Temperature:

Equilibrium constants change when temperature changes; the equilibrium constant really is constant when temperature does not change.

2So₂(g)+o₂(g)←→2So₃(g)

Concentration:

If we change the concentration of the reactants or products in a reaction at equilibrium, the proportions of reactants and products adjust themselves in such a way that Kc does not change.

 Pressure:

Equilibrium constants do not change when pressure changes. For example, if the sulphide dioxide reaction above is performed at 1atm or 10atm then Kp keeps the same value, provided the temperature is 298k.

Another influence of equilibrium constant is catalyst

However, the way the equilibrium constant itself is written changes from reaction to the general rule is, for that we can write as;

Qq+rR←→Ss+Tt

This equation reads q moles of compound Q react with r moles of compound R to give s moles of the compound S and t moles of compound T; the equilibrium constant is given by;

K_c=[S]^s[T]^t/[Q]^q[R]^r

Or

K_p=[P_S]^s[P_T]^t/[P_Q]^q[P_R]^r

The equilibrium constant K_c or K_p is used depending on whether the compounds are in solution or are gases.  Applying my Equilibrium Reaction Law, then it is given;

∆H^ѳ=[∆H^ѳ+∆H^ѳ]/2=T∆S^ѳ-RtlnK_c=T∆S^ѳ-∆G 
∆H^ѳ=[∆H^ѳ+∆H^ѳ]/2=T∆S^ѳ-RtlnK_p=T∆S^ѳ-∆G

PARAMETERS:

∆S be entropy change in the system

∆H be energy available for reaction

T∆S be heat change

∆G be energy of doing work

K be equilibrium constant

EXAMPLE:

Given the reaction at temperature 573k

∆S^ѳ =175.1JK^-1mol^-1

ZnCo₃(s)←→Zno(s)+Co₂(g)

Estimate ∆H^ѳ, ∆G^ѳ and K

At equilibrium, ∆G= T∆S-∆H=0

  

POSTULATE: POINT HYDROGEN SPECTRUM

For all the principal quantum number of the higher level, m, and n the principal quantum number of the lower level. The wave length of principal quantum number of the higher level ,m, plus the wave length of the principal quantum number of the lower level, n, all divided by two is proportionate to the wave length  λ in order of differences to infinity.

λ=[ℓ_aλ_*+ℓ_sλ_*]/2

for each ℓ_a=2, 3, 4,……correspondence ℓ_s=0, -1, -2,…..l..

λ_*=n²/2R_H=hcn²/2k

λ_*=-m²/2R_H=-hcm²/2k

λ be wave length

h be plank constant

C be speed of light