One mole of `Cl_(2(g))` which may be assumed to obey the ideal gas law, initially at 300 K and `1.01325 xx 10^(7)` Pa, is expanded against a constant exteranl pressure of `1.01325 xx 10^(5)` Pa to a final pressure of `1.01325 xx 10^(6)` Pa . As a result of the expansion , the gas cooled to a temperature of 239 K (which is the normal boiling point of `Cl_(2)`) and 0.100 mol of `Cl_(2)` condensed. The enthalpy of vaporization of `Cl_(2(l))` is 20.42 `KJ "mol"^(-1)` at the normal volume is `C_(v)=28.66 JK^(-1) "mol"^(-1)` and the density of `Cl_(2(l))` is 1.56 g `cm^(-1)` (at 239 K) . Assume that the molar heat capacity at constant pressure for `Cl_(2(g))` is `C_(p)=C_(v)+R.` `(1 atm = 101325 xx 10^(5) Pa , R=8.314510 Jk^(-1) "mol"^(-1) = 0.0820584 L atm K^(-1)"mol"^(-1))` (i) Either draw a complete molecular orbital energy diagram or write the complete electronic configuration of `Cl_(2)` Predict the bond order of `Cl_(2)` and thus whether this molecules will be diamagnetic , ferromagnetic ,or paramagnetic. (ii) For the changes decribed above, calculate the change in the internal energy `(DeltaE)` and the change in the entropy `(DeltaS_("sys"))` of the system.

One mole of `Cl_(2(g))` which may be assumed to obey the ideal gas law

1.	One mole of `Cl_(2(g))` which may be assumed to obey the ideal gas law, initially at 300 K and `1.01325 xx 10^(7)` Pa, is expanded against a constant exteranl pressure of `1.01325 xx 10^(5)` Pa to a final pressure of `1.01325 xx 10^(6)` Pa . As a result of the expansion , the gas cooled to a temperature of 239 K (which is the normal boiling point of `Cl_(2)`) and 0.100 mol of `Cl_(2)` condensed. The enthalpy of vaporization of `Cl_(2(l))` is 20.42 `KJ "mol"^(-1)` at the normal volume is `C_(v)=28.66 JK^(-1) "mol"^(-1)` and the density of `Cl_(2(l))` is 1.56 g `cm^(-1)` (at 239 K) . Assume that the molar heat capacity at constant pressure for `Cl_(2(g))` is `C_(p)=C_(v)+R.` `(1 atm = 101325 xx 10^(5) Pa , R=8.314510 Jk^(-1) "mol"^(-1) = 0.0820584 L atm K^(-1)"mol"^(-1))` (i) Either draw a complete molecular orbital energy diagram or write the complete electronic configuration of `Cl_(2)` Predict the bond order of `Cl_(2)` and thus whether this molecules will be diamagnetic , ferromagnetic ,or paramagnetic. (ii) For the changes decribed above, calculate the change in the internal energy `(DeltaE)` and the change in the entropy `(DeltaS_("sys"))` of the system.
Answer» Correct Answer - (i) Electronic configuration fo a Ci atom: `1s^(2)2s^(2)2px^(2)2py^(2)2pz^(2)3s^(2)3px^(2)3py^(2)3pz^(1)` Significant atomic orbitals (AO) =1(K) + 4(L) + 4(M) =9AO Number of electrons in these Aos:17 Number of molecular orbitals (MO) equals number of AOs: Thus `2 xx [1(K) + 4(L) + 4(M) ]=18` MOs are present in a `Cl_(2)` molecule In the information of `Cl_(2): 2 xx 17 = 34` electrons to go into the 18 MOs. MO decription of `Cl_(2)` : `1sigma^(2)1sigma^()2sigma^(2)2sigma^()3sigma^(2)1pi^(4)3sigma^(2)4sigma^(2)4sigma^(2)5sigma^(2)2pi^(4)2pi^(4)` or `(KK)(LL)(sigma3s)^(2)(sigma^()3s)^(2)(sigma3p)^(2)(pi3p)^(4)(pi^()3p)^(4)` `(sigma1s)^(2)(sigma^()1s)^(2)(sigma2s)^(2)(sigma^()2s)^(2)(sigma2p_(z))^(2)(p2p_(y))^(2)(p2p_(x))^(2)(p^()2p_(x))^(2)(p^()2p_(y))^(2)(p^()2p_(z))^(2)` `(sigma3s)^(2)(sigma^()3s)^(2)(sigma3p_(z))^(2)(p3p_(x))^(2)(p3p_(y))^(2)(p^()3p_(x))^(2)(p^()3p_(y))^(2)(p^()3p_(z))^(0)` Or `(KK)(LL) (sigma3s)^(2)(sigma3p_(z))^(2)(pi3p_(z))^(2)(pi3p_(x))^(2)(pi3p_(y))^(2)(pi3p_(x))^(2)(pi^()3p_(x))^(2)(sigma^()3p_(y))^(2)(sigma^()2p_(z))^(0)` ``assumption :- bond formation is along the z-axis (equivalent formulae for x or y axes are accepted ) Bond order is given by `(n-n^())//2:` `n=18; n^(**) = 16` (18-16)/2 =1 (1 `sigma` bond , no `pi` bond) The `Cl_(2)` molecules has a bond order of 1. The `Cl_(2)` molecules is diamagnetic since there are no unpaired electrons. (ii) Summary of the changes involved: `Cl_(2)(g) 1"mol" 300K underset("cooled"1.013 xx 10^(7) "Pa" (100 atm))overset(DeltaE_(1))rarr Cl_(2)(g) 1" mol" 239 K underset(1.013 xx 10^(5) "Pa" 239 K (1 atm)) overset(DeltaE_(2))rarr Cl_(2)(I) 0.1 "mol"` The total process is an expanison plus an isobaric change of phase (gas to liquid) and since the internal energy (E) is a function of state, the total change in the internal energy is `DeltaE = DeltaE_(1) + DeltaE_(2)`. Process 1: `DeltaE_(1) = Int nC_(v)dT = 1 xx 28.66 xx 239 - 300=-1748.3 J` Note a) `DeltaE` for a perfect gas is a function only of T b) `C_(v)` is constant c) "-" sign means a loss of energy due to the work needed for expansion of 1 mole of gas. process 2: For convenience , the data were manipulated in atm, equivalent procedure in Pa will require the appropriate conversion factor. From an energetic pint fo view, the liquid formation Process 2 can be split into two separate steps: `dot" "` the vaporization heat loss (decreased internal energy,-) from the system into surrounding (since the process takes place at constant pressure, the heat is equal to the change in the enthalpy) `dot " "` the work done by the surrounding in compressing the system to a smaller volume (increased internal energy, +). Volume of gas which condensed is `V=nRT//P = (0.1 xx 0.0820584 xx 239)//1 = 1.96 dm^(3)` Volume of liquid `Cl_(2): (0.1 xx 2 xx 35.454)//1.56 = 4.54 cm^(3)` `DeltaE_(2)=DeltaH_(2)-int P_("ext") DeltaV(phase change)=DeltaH_(2)-P_("ext") (V_(1)-V_(g))` But `V_(1)` is approximately 0 and can be neglected (ca. 4.5` cm^(3)` liquid volume vs. ca. 17.6 `dm^(3)`; ca 0.03% error) `DeltaE_(2) =(0.1)(-DeltaH_("vap")) + P_("ext")Vg` `=0.1 xx (.^(-)C 20420)+ (1 xx 1.96 L ) xx 101.325 J dm^(-3) atm^(-1) = -2042.0 + 198.5 =-1843.5` `DeltaE=DeltaE_(1) + DeltaE_(2) =-1748.3 + -1843.5)=-3591.8` Entropy S is a function of two variables of state. Since in Process 1 the known variables are T and `P_(1)` expression of S is chosen as S(T,P). `Delta S_("sys") = DeltaS_(1) + Deltas_(2)` and `overset(-)C_(p)=overset(-)C_(v) + R =28.66 + 8.314 = 36.97 J K^(-1) "mol"^(-1)` `DeltaS_(1) nC_(p) "In" (T_(2))/(T_(1)) - nR "In" (P_(2))/(P_(1)) = 1.0 xx 36.97 "in" (239)/(300) - 8.314 "in" (1)/(100) ==-8.40 + 38.29` `=29.89 J K^(-1)` For the phase transition (constant temperature) , by definition `DeltaS_(2) = Q//T` Since the pressure is constant in this case ,` Qp//T = Qp//T = DeltaH//T` `DeltaS_(2) = (DeltaH_(2))/(T) = (0.1 xx (-20420))/(239) =-8.54 JK^(-1)` `DeltaS_("sys") =29389 - 8.54 = 21.35 JHK^(-1)`

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