Question

What is Delta G in kJ for the reaction: Cl_2(g)+ 2 Br^- rightarrow Br_2(l) + 2 Cl^- if [Br^-] = 0.100, [Cl^-] = 0.50, and p(Cl_2) = 1.00 atm? The temperature 25 degree C.

Answer 1

Ans. Cl₂(g) + 2 Br^-(aq) = Br₂(l) + 2 Cl^-(aq)

Equilibrium constant of the reaction, K = ([Br₂] [Cl^-]²) / ([Cl₂] [Br^-]²)

The concentration of pure chemical species is taken to be unity.

So, K = [Cl^-]² / [Br^-]² = (0.50)² / (0.10)² = 25

Now,

At equilibrium,

Using the equation dG = dG^0’ + RT lnK              - equation 1

            Where, dG = calculated/ experimental free energy change = ?

                                    dG^0’ = standard/ theoretical free energy change                                                            R = 0.0083146 kJ mol^-1K^-1

                                    T = temperature in kelvin = (⁰C + 273.15) K

                                    K = equilibrium constant under given condition.

We have, dG⁰ for the reaction = - 56.35 kJ/mol

Putting the values in equation 1-

            dG = - 56.35 kJ mol^-1+ (0.0083146 kJ mol^-1 K^-1) x 298.15 K x (ln 25)

            Or, dG= - 56.35 kJ mol^-1 + 2.47899799 kJ mol^-1 x (3.219)

            Or, dG= - 56.35 kJ mol^-1 + 7.98 kJ mol^-1

            Or, dG = - 48.37 kJ mol^-1

Therefore, dG of the reaction = - 48.37 kJ mol^-1

Note: The value of dG shall be in terms of kJ/ mol. Please use the dG⁰ mentioned in your textbook top get the dG value closer to -48 kJ/mol or round off -48.33 kJ/mol to -48 kJ/mol.