My question is ..can anyone explain how to do problem #2? I feel lost. :( And could you also explain why? Problem#1, is that a regular dG = dH - TdS problem? Then using Go = -RTlnK to find K? Im just afraid if there are other seps to it..
1. The combustion of butane has the following balanced equation:
2 C4H10(g) + 13 O2(g) ----> 8 CO2(g) + 10 H2O(l)
a) Calculate the value of dG for the combustion of C4H10(g).
b) Calculate the equilibrium constant K at 500°C for the combustion of butane.
2. The following reaction is nonspontaneous under standard conditions:
Cd(s) + Fe^2+ ---> Fe(s) + Cd^2+
a) Calculate the temperature(s) at which this reaction would be spontaneous. The dS° of this reaction is 40.0 J/K. (Use at least 3 sig figs in your calculations).
b) Calculate the concentration of Cd^2+ needed for a cell that contains 1.00 M Fe^2+ to be spontaneous.
c) Explain why a higher/lower concentration of cadmium ion makes the reaction spontaneous. Your answer should use a concept about/related to equilibrium.
For #1, I suspect they expect you to calculate dGo which you get from dGo = dHo - TdSo from tables. However, most tables have dGo already tabulated and you need not go through the dH-TdS part.
Then dGo = -RTlnK should get it.
#2.
I'll look at 2 later.
To solve problem #2, let's break it down step by step:
a) To determine the temperature at which this reaction would be spontaneous, we need to use the equation ΔG = ΔH - TΔS, where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.
Given that the reaction is nonspontaneous, ΔG > 0. Therefore, we can rearrange the equation:
ΔG = ΔH - TΔS
ΔH = TΔS (since ΔG > 0)
To find the temperature, T, at which the reaction becomes spontaneous, we solve for T:
T = ΔH/ΔS
Substitute the values given in the problem:
T = 0 / 40.0 J/K (ΔH is assumed to be zero since the reaction is nonspontaneous)
Therefore, T = 0 K. This means that the reaction would only be spontaneous at absolute zero temperature.
b) To calculate the concentration of Cd^2+ needed for a cell that contains 1.00 M Fe^2+ to be spontaneous, we need to use the Nernst equation:
Ecell = E°cell - (0.0592V / n)log(Q)
Where Ecell is the cell potential, E°cell is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient.
Since the reaction is spontaneous, we know the cell potential, Ecell, is positive. Therefore, we can rearrange the equation as follows:
E°cell - (0.0592V / n)log(Q) > 0
-log(Q) > -E°cell * n/0.0592V
log(Q) < E°cell * n/0.0592V
The reaction quotient, Q, is calculated using the concentrations of the products and reactants:
Q = [Fe][Cd^2+]/([Cd^2+][Fe^2+])
To make the reaction spontaneous, we need to increase the concentration of Cd^2+ until log(Q) is lower than the right side of the inequality.
c) A higher concentration of cadmium ion (Cd^2+) makes the reaction spontaneous because it increases the value of the reaction quotient, Q. As Q becomes larger, the log(Q) term in the Nernst equation becomes more negative, making the inequality Q < E°cell * n/0.0592V true. This shift toward higher concentration of Cd^2+ favors the forward reaction, making it spontaneous.