The fermentation of sugar (C6H12O6) to form ethanol (C2H5OH) and carbon dioxide (CO2) occurs according to the following equation:
C6H12O6(aq) ¡ú 2 C2H5OH(aq) + 2 CO2(g)
kJ mol¨C1
¦¤Hf (C6H12O6(s))
is ¨C1275
¦¤Hf (C2H5OH(l))
is ¨C277
¦¤Hbond (C=O)
799
(a) Assuming the heat of dissolution of C6H12O6(s) and C2H5OH(l) in water are negligible, calculate an approximate value for the enthalpy for the above reaction.
(b) An approximate value for the entropy change, ¦¤S, for this reaction is +376 J K¨C1 mol¨C1. Comment on the sign and magnitude of this value and calculate values for the Gibbs free energy, ¦¤G, and the equilibrium constant, K at 298 K.
(c) State under what conditions this reaction is spontaneous.
a) Well, if we take a look at the equation, we can see that 1 mole of C6H12O6 produces 2 moles of C2H5OH and 2 moles of CO2. So let's do some quick math - if the enthalpies of formation for C6H12O6 and C2H5OH are -1275 kJ/mol and -277 kJ/mol respectively, we can calculate the enthalpy change for the reaction.
ΔH = [2 × ΔHf(C2H5OH)] + [2 × ΔHf(CO2)] - ΔHf(C6H12O6)
= [2 × (-277 kJ/mol)] + [2 × 0 kJ/mol] - (-1275 kJ/mol)
= -554 kJ/mol + 0 kJ/mol + 1275 kJ/mol
= 721 kJ/mol
So the approximate value for the enthalpy change of the reaction is 721 kJ/mol.
b) Now let's move on to the entropy change. The provided value is ΔS = +376 J K^-1 mol^-1. The positive sign indicates an increase in disorder, which makes sense since we are going from a solid (C6H12O6) to two aqueous molecules (C2H5OH) and a gas (CO2). As for the magnitude, it's of moderate size.
Now, to calculate the Gibbs free energy change (ΔG), we can use the equation:
ΔG = ΔH - TΔS
At 298 K, we substitute the values:
ΔG = 721 kJ/mol - (298 K * 0.376 kJ K^-1 mol^-1)
ΔG ≈ 721 kJ/mol - 111 kJ/mol
ΔG ≈ 610 kJ/mol
For the equilibrium constant (K), we can use the relationship:
ΔG = -RT ln K
Substituting the values:
610 kJ/mol = -(8.314 J K^-1 mol^-1)(298 K) ln K
Solving for K:
ln K = -610 kJ/mol / (-(8.314 J K^-1 mol^-1)(298 K))
K ≈ e^(-610 / (8.314 * 298))
And we get the approximate value for the equilibrium constant.
c) The reaction is spontaneous when ΔG is negative. That means it will occur naturally and without any encouragement. So, for this reaction to be spontaneous, the temperature must be lower than the temperature at which ΔG becomes positive.
(a) To calculate the approximate value for the enthalpy change (ΔH) of the reaction, you need to consider the enthalpy of formation (ΔHf) values for the reactants and products.
The balanced reaction is:
C6H12O6(aq) → 2 C2H5OH(aq) + 2 CO2(g)
ΔHf for C6H12O6(s) = -1275 kJ/mol
ΔHf for C2H5OH(l) = -277 kJ/mol
When balancing the reaction, you find that the coefficients in the balanced equation represent moles of each compound reacting. So, the enthalpy change can be calculated using the ΔHf values:
ΔH = (2 × ΔHf[C2H5OH(l)] + 2 × ΔHf[CO2(g)]) - ΔHf[C6H12O6(s)]
Plugging in the values, we get:
ΔH = (2 × (-277 kJ/mol) + 2 × 0 kJ/mol) - (-1275 kJ/mol)
ΔH = (-554 kJ/mol) + 1275 kJ/mol
ΔH ≈ +721 kJ/mol
Therefore, the approximate enthalpy change for the reaction is around +721 kJ/mol.
(b) The entropy change (ΔS) for the reaction is given as +376 J K^(-1) mol^(-1). A positive value for ΔS indicates an increase in entropy, meaning that the reaction leads to more disorder in the system.
To calculate the Gibbs free energy change (ΔG) at 298 K, you can use the equation:
ΔG = ΔH - TΔS
Plugging in the values:
ΔG = (+721 kJ/mol) - (298 K × (376 J K^(-1) mol^(-1) / 1000))
ΔG ≈ +721 kJ/mol - 112.048 kJ/mol
ΔG ≈ +608 kJ/mol
The positive ΔG value indicates that the reaction is non-spontaneous under standard conditions (298 K and 1 atm pressure).
The equilibrium constant (K) can be calculated using the equation:
ΔG = -RT ln(K)
Plugging in the values:
+608 kJ/mol = -((8.314 J K^(-1) mol^(-1)) × 298 K) ln(K)
ln(K) ≈ -(+608 kJ/mol) / ((8.314 J K^(-1) mol^(-1)) × 298 K)
Simplifying, we get:
ln(K) ≈ -0.2504
K ≈ e^(-0.2504)
K ≈ 0.779
Therefore, the equilibrium constant (K) for this reaction at 298 K is approximately 0.779.
(c) The reaction is spontaneous under the following conditions:
1. At temperatures higher than 298 K: Since ΔG is positive at 298 K, an increase in temperature will decrease ΔG, making the reaction spontaneous.
2. With a decrease in pressure: Since CO2(g) is a product gas, reducing the pressure will shift the equilibrium towards more CO2(g) formation, making the reaction spontaneous.
3. With a decrease in the concentration of C2H5OH(aq): Lowering the concentration of the product C2H5OH(aq) will shift the equilibrium towards more C2H5OH(aq) formation, making the reaction spontaneous.
4. With an increase in the concentration of C6H12O6(aq): Increasing the concentration of the reactant C6H12O6(aq) will shift the equilibrium towards more C2H5OH(aq) and CO2(g) formation, making the reaction spontaneous.
Note: These conditions apply assuming the reaction is taking place at 298 K.
(a) To calculate the enthalpy change (∆H) for the reaction, we can use the standard enthalpies of formation (∆Hf) of the products and reactants involved. The standard enthalpy change (∆H°) can be calculated using the following formula:
∆H° = ∑(∆Hf° products) - ∑(∆Hf° reactants)
In this case, we need to consider the enthalpies of formation for C6H12O6(s), C2H5OH(l), and CO2(g):
∆Hf° (C6H12O6(s)) = -1275 kJ/mol
∆Hf° (C2H5OH(l)) = -277 kJ/mol
∆Hf° (CO2(g)) = 0 kJ/mol (since it is the standard state)
Since there is one mole of C6H12O6 forming 2 moles of C2H5OH and 2 moles of CO2, we multiply the ∆Hf° of C2H5OH(l) and CO2(g) by 2:
∆H° = [2 * ∆Hf° (C2H5OH(l))] + [2 * ∆Hf° (CO2(g))] - ∆Hf° (C6H12O6(s))
= [2 * (-277 kJ/mol)] + [2 * (0 kJ/mol)] - (-1275 kJ/mol)
= -554 kJ/mol + 0 kJ/mol + 1275 kJ/mol
= 721 kJ/mol
Therefore, the approximate enthalpy change (∆H) for the given reaction is +721 kJ/mol.
(b) The sign and magnitude of the entropy change (∆S) value can give us insights into the spontaneity of the reaction. A positive ∆S value indicates an increase in entropy while a negative ∆S value indicates a decrease in entropy.
In this case, ∆S is given as +376 J K^-1 mol^-1, which is positive. This suggests that the reaction leads to an increase in entropy.
To calculate the Gibbs free energy change (∆G°) for the reaction, we can use the equation:
∆G° = ∆H° - T∆S°
where T is the temperature in Kelvin (298 K in this case).
∆G° = 721 kJ/mol - (298 K * (376 J K^-1 mol^-1 / 1000 J/kJ))
= 721 kJ/mol - (298 K * 0.376 kJ/mol)
= 721 kJ/mol - 111.848 kJ/mol
= 609.152 kJ/mol
Therefore, the ∆G° value for the given reaction at 298 K is approximately 609.152 kJ/mol.
To calculate the equilibrium constant (K) at 298 K, we can use the following equation:
∆G° = -RT ln(K)
where R is the gas constant (8.314 J K^-1 mol^-1).
Rearranging the equation, we have:
K = e^(-∆G° / (RT))
K = e^(-609.152 kJ/mol / ((8.314 J K^-1 mol^-1) * 298 K))
= e^(-203090.768 / 2475.572)
= e^-82.048
Using the exponential function on a calculator, we find that K is approximately 1.391 * 10^-36.
(c) This reaction is spontaneous under the following conditions:
1. At temperatures where the ∆H° is positive (in this case, ∆H° = +721 kJ/mol).
2. At temperatures where the ∆S° is positive (in this case, ∆S° = +376 J K^-1 mol^-1).
3. At temperatures where ∆G° is negative (in this case, ∆G° = -609.152 kJ/mol).
Since the ∆H° is positive and the ∆S° is positive, the reaction has a favorable change in entropy. Additionally, the negative value of ∆G° indicates that the reaction is spontaneous.