Calculate the standard entropy, ΔS°rxn, of the following reaction at 25.0 °C using the data in this table. The standard enthalpy of the reaction, ΔH°rxn, is –44.2 kJ·mol–1.
C2H4(G)+H20 ---> C5H5OH
ΔS°rxn= ______ J*K^-1*mol^-1
Then, calculate the standard Gibbs free energy of the reaction, ΔG°rxn.
ΔG°rxn.=_______ kJ*mol^-1
∆Hf° ∆Gf° S°
C2H4(g) 52.4 68.4 219.3
H2O(l) –285.8 –237.1 70.0
C2H5OH(l) –277.6 –174.8 160.7
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Hint
The standard entropy of the reaction, ΔS°rxn, is equal to the sum of the standard entropies, ΔS°, of the products multiplied by their coefficients from the balanced equation, n, minus the sum of the ΔS° of the reactants multiplied by their coefficients, m.
ΔS°rxn=Products - Reactants
Then,use the values of ΔH°rxn and ΔS°rxn to calculate the standard Gibbs free energy of the reaction, ΔG°rxn.
-128.6
Sure, let's calculate the standard entropy, ΔS°rxn, of the reaction first.
The balanced equation is: C2H4(g) + H2O(l) → C2H5OH(l)
ΔS°rxn = (n * ΔS° of products) - (m * ΔS° of reactants)
n and m are the coefficients in the balanced equation.
n = 1 (coefficient of C2H5OH)
m = 1 (coefficient of C2H4)
ΔS°rxn = (1 * ΔS° of C2H5OH) - (1 * ΔS° of C2H4)
Now, let's look up the values of ΔS° for C2H5OH and C2H4 in the table:
ΔS° of C2H5OH = 160.7 J·K^-1·mol^-1
ΔS° of C2H4 = 219.3 J·K^-1·mol^-1
ΔS°rxn = (1 * 160.7) - (1 * 219.3) = -58.6 J·K^-1·mol^-1
Now, let's calculate the standard Gibbs free energy, ΔG°rxn, using the given ΔH°rxn and ΔS°rxn.
ΔG°rxn = ΔH°rxn - TΔS°rxn
T = temperature in Kelvin (25.0 °C = 298.15 K)
ΔG°rxn = -44.2 kJ·mol^-1 - (298.15 K * (-58.6 J·K^-1·mol^-1))
ΔG°rxn = -44.2 kJ·mol^-1 + 18.2 kJ·mol^-1
ΔG°rxn = -26.0 kJ·mol^-1
So, ΔS°rxn = -58.6 J·K^-1·mol^-1 and ΔG°rxn = -26.0 kJ·mol^-1.
To calculate the standard entropy, ΔS°rxn, of the reaction, we need to use the formula:
ΔS°rxn = Σ(n * ΔS° products) - Σ(m * ΔS° reactants)
Looking at the balanced equation for the reaction:
C2H4(g) + H2O(l) ---> C2H5OH(l)
The coefficients of the products and reactants are:
n = 1 (coefficient of C2H5OH(l))
m = 1 (coefficient of C2H4(g)) + 1 (coefficient of H2O(l))
Now, let's calculate ΔS°rxn:
ΔS°rxn = (1 * ΔS° C2H5OH) - ((1 * ΔS° C2H4) + (1 * ΔS° H2O))
Using the data from the table:
ΔS° C2H5OH = 160.7 J*K^-1*mol^-1
ΔS° C2H4 = 219.3 J*K^-1*mol^-1
ΔS° H2O = 70.0 J*K^-1*mol^-1
ΔS°rxn = (1 * 160.7) - ((1 * 219.3) + (1 * 70.0))
ΔS°rxn = 160.7 - (219.3 + 70.0)
ΔS°rxn = 160.7 - 289.3
ΔS°rxn = -128.6 J*K^-1*mol^-1
Therefore, the standard entropy of the reaction, ΔS°rxn, is -128.6 J*K^-1*mol^-1.
Now, let's calculate the standard Gibbs free energy, ΔG°rxn, using the formula:
ΔG°rxn = ΔH°rxn - T * ΔS°rxn
Given that ΔH°rxn = -44.2 kJ·mol^-1 and converting from kJ to J:
ΔH°rxn = -44.2 × 1000 J·mol^-1
ΔS°rxn = -128.6 J·K^-1·mol^-1
And given that the temperature, T = 25.0 °C = 298.15 K
ΔG°rxn = (-44.2 × 1000) - (298.15 * (-128.6))
ΔG°rxn = -44200 - (-38421.679)
ΔG°rxn ≈ -5778.321 kJ·mol^-1
Therefore, the standard Gibbs free energy of the reaction, ΔG°rxn, is approximately -5778.321 kJ·mol^-1.
To calculate the standard entropy, ΔS°rxn of the reaction, use the following steps:
1. Write down the balanced equation for the reaction:
C2H4(G) + H20 => C5H5OH
2. Determine the coefficients of the reactants and products from the balanced equation:
Reactants: C2H4(G) has a coefficient of 1 and H20 has a coefficient of 1.
Products: C5H5OH has a coefficient of 1.
3. Look up the standard entropy values (S°) for the reactants and products in the given table:
S° for C2H4(G) is 219.3 J*K^-1*mol^-1.
S° for H2O (l) is 70.0 J*K^-1*mol^-1.
S° for C5H5OH (l) is 160.7 J*K^-1*mol^-1.
4. Calculate the ΔS°rxn using the formula:
ΔS°rxn = (Sum of Products × Coefficient) - (Sum of Reactants × Coefficient)
ΔS°rxn = (160.7 × 1) - (219.3 × 1)
ΔS°rxn = -58.6 J*K^-1*mol^-1
Therefore, ΔS°rxn = -58.6 J*K^-1*mol^-1
To calculate the standard Gibbs free energy, ΔG°rxn, of the reaction, use the ΔH°rxn and ΔS°rxn values:
1. Convert the standard enthalpy change, ΔH°rxn, from kJ/mol to J/mol:
ΔH°rxn = -44.2 × 10^3 J/mol
2. Use the formula:
ΔG°rxn = ΔH°rxn - (T × ΔS°rxn)
where T is the temperature in Kelvin (25.0 °C = 298 K) and ΔS°rxn is the standard entropy change.
3. Calculate ΔG°rxn using the given values:
ΔG°rxn = (-44.2 × 10^3) - (298 × (-58.6))
ΔG°rxn = -44.2 × 10^3 + 17.4 × 10^3
ΔG°rxn = -26.8 × 10^3 J/mol
Finally, ΔG°rxn = -26.8 kJ/mol.
I can't read the values you have in the table because of the problem with spacing on this forum but here is what you do.
dSo = (n*dS products) - (n*dS reactants)
dH you have but it can be done the same way.
After you have dSo rxn and dHo rxn, then'
dGorxn - dH - TdS.