Question

I need some help for the calculations on Experiment # 22: Enthalpy, Entropy, and Free energy. For the reaction Zn(s) + CuSO₄(aq) ⃡ ZnSO₄(aq) + Cu(s), I need to:

a. Calculate the values of the ΔS and the ΔH, showing units

b. Calculate values of the change in G from equations 1 and 2, showing units.

Here are my values:

Temperature (K)	Voltage (V), (E0)
333.15	1.120
328.15	1.117
323.15	1.112
318.15	1.109
313.15	1.106
308.15	1.105
303.15	1.104
298.15	1.103

And here are the equations:

Equation 1: ΔG = ΔH - TΔS

Equation 2: ΔG = -nFE

Answer 1

So we have the given values of E(cell potential) at various temperature. Now in order to calculate G,   H & S from the above data, we have to make use of both the equations 1 & 2.

a) Let's consider the first two temperatures, T₁=333.15 K & T₂=328.15 K.

At T₁ ,we haveG = -nFE = - 2 x 96485.33 C/mole x 1.12 volts = -216127.14 J/mole = -216.127 kJ/mole

-216127.14 = H - 333.15 x S ............. (3)

At T₂ , we have G = -2 x 96485.33 C/mole x 1.117 volts = -215548.23 J/mol = -215.548 kJ/mol

-215548.23 =  H - 328.15 x S ................ (4)

Combining equation 3 & 4 we get ,

H = Enthalpy = -177555.023 J /mole = -177.55 kJ/mole

S = Entropy = 115.78 J / mol-K

b) After calculating  H &  S, let's keep these two quantities constant and use the temperature as the variable. Using equation 1 & 2 , we would find the difference in values of  G.

At T₃ = 323.15 K , we have  G from eq 1 =  H - T  S = -177555.023 - 323.15 x 115.78 = -177555.023 - 37414.307 = -214969.33 J/mol = -214.97 kJ/mol

At T₃ using equation 2, we get  G = - 2 x 96485.33 x 1.112 J/mole = -214583.374 J/mole = -214.58 kJ/mole

Now, the difference in Gibbs energy between the two equations = -214583.374 - (-214969.33) = 385.956 J/ mole.