ORTEP diagram for CCl₄

Q2006. Concentration of a coloured complex in solution can be calculated from absorption measurements using the Beer-Lambert's Law A = ecL  where A is the absorbance, e the molar absorption coefficient, L is the cell length and c is the molar concentration of the coloured solute. Thus we get c = A/(eL) giving the concentration.

To let the computer read a set of input values for A, e and l for absorbance studies of Fe - 1,10-phenanthroline complex so as to calculate the corresponding molar concentrations, we may use the following Fortran program:

C      Calculation of concentration from absorbance measurement
C      (for N sets of input data values)
       WRITE(*,*) 'Enter the number of data sets:'
       READ(*,1) N
  1    FORMAT(I4)
       DO 7 I = 1, N, 1
         WRITE(*,2)
  2      FORMAT(/1X,'Enter Absorbance, Abs. Coeff., Cell-Length:')
         READ(*,*) Abs, Coeff, CLeng
         Conc = Abs/(Coeff*CLeng)
         WRITE(*,3)
  3      FORMAT(/1X,'Calculated Concentration is:',1X,F11.8)
  7    CONTINUE
       STOP
       END 

Dear students, in format statements I4 means an upto 4-digit integer, / leads to next line, 1X creates a blank space, and F11.8 keeps 11 spaces for the real-number output with 8 places after decimal. Every DO statement must contain a line identifier (such as 7 here) with which line number the corresponding CONTINUE statement is identified.

Q2005. The progress of the gaseous reaction 2N₂O₅ → 4NO₂ + O₂ is best monitored by monitoring the change of total pressure (P) for the reaction mixture as a function of time t, with the reaction occurring at a fixed-volume container at constant temperature. Writing the reaction as N₂O₅ → 2NO₂ + 0.5 O₂  where all the reactants and products are gaseous, we have (assuming PV = nRT) P  α  {(n_o – x) + 2x + 0.5x} with initial pressure P_o  α  n_o (where x is the no. of moles of N₂O₅ decomposed) which gives (P – P_o) = (3/2)Cx while P_o  =  Cn_o where C is a constant. This gives (Cn_o – Cx) = {P_o – 2(P – P_o)/3} = (5P_o – 2P)/3, i.e., [N₂O₅] or (n_o – x) is proportional to (5P_o – 2P)/3 or to (5P_o – 2P), while when n = n_o , P = P_o., so that then (5P_o – 2P) = 3P_o. This means that [N₂O₅]_o / [N₂O₅] = 3P_o/(5P_o – 2P). As the reaction is a first order one obeying the rate law rate = k₁[N₂O₅], so this means that (as k₁ = (2.303/t) log [N₂O₅]_o / [N₂O₅] ) that k₁ =  (2.303/t) log {3P_o/(5P_o – 2P)}. Prof. Raman mistakenly thinks that partial pressure P_p of N₂O₅ could be monitored with time (how, sir!), while obviously [N₂O₅]  α  P_p [we are unlikely to detect his mistake]. However, if there the partial pressure P_p of N₂O₅ could be monitored with time t, P_p being proportional to concentration [N₂O₅] of N₂O₅, we directly get (as k₁ = (2.303/t) log [N₂O₅]_o / [N₂O₅] ) that k₁ =  (2.303/t) log {P_po/P_p). The easiest way to calculate the rate constant (without employing least square fitting etc.) is so to calculate (with above relation) values of rate constant k₁ for every (time, pressure) data-point and then to calculate the average. The following FORTRAN program does this work:

C      Calculation of 1st-order reaction rate constant from partial pressures
C      (for N number of non-initial input data points)
       WRITE(*,11)
 11    FORMAT(/1X,'Enter initial partial pressure Pp0 of reactant:')
       READ(*,*) Pp0
       WRITE(*,12)
 12    FORMAT(/1X, 'Enter number of beyond-initial (time,Pp) data points:')
       READ(*,*) N
       SUMk1 = 0.0
       DO 7 I = 1, N, 1
         WRITE(*,22)
 22      FORMAT(/1X,'Enter time t and reactant partial pressure Pp:')
         READ(*,*) t,Pp
         Factr = Pp0/Pp
         SUMk1 = SUMk1 + (2.303/t)*ALOG10{Factr)
  7    CONTINUE
       RtCnst = SUMk1/FLOAT(N)
       WRITE(*,30) RtCnst
 30    FORMAT(/1X,'Rate Constant:',1X,F12.7)
       STOP
       END   

The method of linear least square fitting is applicable for a set of experimental points (x,y) regarding a relation between the function y = f(x) and its argument (i.e., variable) x, where there is a underlying linear relation (of the form y = mx + c) between the var5iables y and x. In this method, the vertical distance d_i = {y_i – (mx_i + c)} between the experimental point (x_i,y_i) and the linear-fitted point (x,mx+c) is considered for an arbitrary i-th point. The sum of the squares of such distances is a function of the fitting parameters m & c, and this sum is minimised as per differential calculus to get the optimum values of m & c. The m and c values thus come out to be definite functions of the set of (x_i,y_i) values, given by relations m = ?/?? and c = ???/.??. Using a computer, the expressions for m and c could be easily calculated for any given set of data, as per the following FORTRAN program:

PROGRAM....
Dear students, may keep Read statements unformatted -- this is advisable for generality. May change Write statements as given in above examples - both pre-Read Write statements and result-displaying ones (see above e.g.).

2006.(i) (a**3) + 3*a*b(a+b) + (b**3)     (ii) a / SQRT((ah**2)+(ak**2)+(al**2))

2005. We should use statements amiu = (em1*em2)/(em1+em2)   and (a+b)**2 = (a**2) + 2*a*b + (b**2)

2005. Say that we should use a Fortran program for solving quadratic equation. Write that program codes (change the Write statements there to formatted ones as per above examples). Then say that we need to prepare the executable (.exe) program file as per the FTN77, FCG77, LINK86/F procedure, then run this executable program and input the coefficient a of E² as 0.75, coefficient b of E as 0.35 and the constant term c as –0.023. The solution for E will then be displayed in the computer. 

2006: FORTRAN program for delocalization energy of conjugated linear polyenes (e.g., butadiene) and conjugated cyclic polyenes (e.g., benzene) as per Huckel MO theory:
As per the Huckel MO theory, we find that for a conjugated organic molecule containing N_C carbon atoms (for butadiene, N = 4) and thus N_C p-electrons, there are N_C 2pp Huckel-MOs numbered k = 1, 2,..... N_C (see B. K. Sen, Unit 5.17) with energies e_k as (i) For conjugated linear polyenes e_k = a + 2b cos{kp/(N_C+1)} and (ii) For conjugated cyclic polyenes
e_m = a + 2b cos(mp/N_C) where m = 0,+1,–1,... N_C/2 for even N_C (e.g., for benzene with N_C = 6) and
m = 0,+1,–1,... (N_C–1)/2,–(N_C–1)/2 for odd N_C (e.g., for cyclopentadienyl radical, C₅H₅' with N_C = 5)
The Huckel p-electron energy of the molecule is thus E = S_k n_ke_{k ,} where n_k (2, 1, or 0) is the no. of electrons occupying the k-th Huckel MO. For any of the two classes of conjugated polyenes, we have the non-delocalized (i.e., localized) energy E_loc = N(a + b) [this concept considers 1,3 butadiene as a combination of just two separate ethene molecules] Thus the delocalization energy E_dl is:
E_dl = S_k n_ke_k – N(a + b) = E – N(a + b) 

To implement this calculation for a conjugated linear polyenes (e.g., butadiene) in a FORTRAN program, we name variable  N_C as NC, k as K, n_k as NK, e_k as EK, a as ALPHA, b as BETA. For such linear conjugated polyenes, we have n_k = 2 for k less than or equal to N_C/2 (N_C is obviously even here), and n_k = 0 for k > N_C/2. Further for butadiene specifically, N_C = 4. So we have the following FORTRAN program for delocalization-energy calculation of butadiene:

C     Calculation of delocalization-energy for butadiene (NC = 4)
      NC = 4
      WRITE(*,2)
 2    FORMAT(/1X,'Enter values of alpha and beta:')
      READ(*,*) ALPHA, BETA
      E = 0.0
      DO 3 K = 1,NC,1
        EK = ALPHA + 2.0*BETA*COS(FLOAT(K)*3.1416/(FLOAT(NC)+1))
        NCHalf = NC/2
        IF (K .LE. NCHalf) THEN
          NK = 2
        ELSE
          NK = 0
        ENDIF
        WRITE(*,*) 'Level No.:',K,' No. of e-:',NK
        E = E + FLOAT(NK)*EK
        WRITE(*,*) 'H-MO Energy is:',EK
        WRITE(*,*) 'Tot. Huckel Energy is:',E
 3    CONTINUE
      ELOC = FLOAT(NC)*(ALPHA+BETA)
      EDL = E - ELOC
      WRITE(*,4) EDL
 4    FORMAT(/1X,'Delocalisation Energy:',1X,F15.7)
      STOP
      END

2006: Whether a given conjugated cyclic organic molecule is aromatic:
Whether a conjugated cyclic organic molecule is aromatic is or not is known by the simple consideration of whether the number of pi-electrons there equals (4n+2), where n is a non-negative integer (the Huckel 4n+2 rule). Thus we need to consider the number of pi-electrons (say, NoPiEl) in the molecule, and starting from n = 0, check whether that number equals 4n+2. So the FORTRAN program is as follows: 

C     Whether a conjugated cyclic organic molecule is aromatic
      WRITE(*,2)
 2    FORMAT(/1X,'Enter the number of pi-electrons in molecule:')
      READ(*,*)NoPiEl
      NArom = 0 
      DO 3 N = 0,200,1
        NTest = 4*N + 2
        IF (NTest .EQ. NoPiEl) THEN
          NArom = 1
        ENDIF
 3    CONTINUE
      IF (NArom .EQ. 1) THEN
        WRITE(*,4)
 4      FORMAT(/1X,'The molecule is aromatic')
      ELSE
        WRITE(*,5)
 5      FORMAT(/1X,'The molecule is not aromatic')
      ENDIF
      STOP
      END