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Using E-mail to Discuss Chemistry 

E-mail me your final answer to the problem sets.  If the question is a calculation, send the answer and a step-by-step description of how you arrived at the answer (you need not try to type in math).   The following two examples illustrate appropriate answers and both show how to set chemistry in e-mail without the use of superscripts or subscripts. 

Example 1 
Example 2 

Example 1 

Under certain conditions, the equilibrium constant for the decomposition of PCl₅(g) into PCl₃(g) and Cl₂(g) is 0.0211 mol L^-1. What are the equilibrium concentrations of PCl₅, PCl₃, and Cl₂ if the initial concentration of PCl₅ was 1.00 M?

Solution for Example 1 (Either repeat the question or indicate its number.) 

Step 1.  Determine the direction the reaction shifts.  Initially we have no products so Q = 0 and the reaction will shift to the right. 

Step 2.  Determine the relative changes needed to reach equilibrium then write the equilibrium concentrations in terms of these changes.  We represent the increase in concentration of PCl3 by the symbol X.  The other changes may be written in terms of X.  ([ ]i = initial concentration; [ ]eq = concentration at equilibrium) 

                 PCl5    =    PCl3  +    Cl2 
[ ]i, M          1.00          0          0 
Change, M        -X            X          X 
[ ]eq M       1.00 + (-X)      X          X

    X = 0.135 M

    [PCl5] = 1.00 - 0.135 = 0.86 M 
    [PCl3] = X = 0.135 M 
     [Cl2] = X = 0.135 M

ClF₃ can be prepared by the reaction represented by the equation above. For ClF₃ the standard enthalpy of formation, DH_f, is - 163.2 kilojoules/mole and the standard free energy of formation, DG_f, is - 123.0 kilojoules/mole. 

(a) Calculate the value of the equilibrium constant for the reaction at 298 K. 

(b) Calculate the standard entropy change, DS, for the reaction at 298 K. 

(c) If ClF₃ were produced as a liquid rather than as a gas, how would the sign and magnitude of DS for the reaction be affected? Explain. 

(d) at 298 K the absolute entropies of Cl₂(g) and ClF₃(g) are 222.96 joules per mole-Kelvin and 281.50 joules per mole-Kelvin, respectively. 

(i) Account for the larger entropy of ClF₃(g) relative to that of Cl₂(g). 
(ii) Calculate the value of the absolute entropy of F₂(g) at 298 K. 

(a) rearranging [delta]G^o = - RT ln K gives K. 

K = 3.72 x 10^21 

[delta]S^o = - 270 J/K

[delta]Srxn = [sum]Sproducts - [sum]Sreactants 

S(F2) = 203 J mol^-1  K^-1