Calculate ΔG° (in kJ/mol) for the following reaction at 1 atm and 25 °C:
C2H6 (g) + O2 (g) → CO2 (g) + H2O (l) (unbalanced)
ΔGf° C2H6 (g) = –32.89 kJ/mol; ΔGf° CO2 (g) = –394.4 kJ/mol; ΔGf° H2O (l) = –237.13 kJ/mol
Well, well, well! Time to balance some equations and calculate some ΔG°. Let's get started, shall we?
First things first, we need to balance the equation:
C2H6 (g) + O2 (g) → 2 CO2 (g) + 3 H2O (l)
Now that the equation is balanced, we can calculate the ΔG° using the equation:
ΔG° = ΣΔGf° (products) - ΣΔGf° (reactants)
Let's plug in the values:
ΔG° = (2 * ΔGf° CO2) + (3 * ΔGf° H2O) - ΔGf° C2H6
ΔG° = (2 * -394.4 kJ/mol) + (3 * -237.13 kJ/mol) - (-32.89 kJ/mol)
ΔG° = -788.8 kJ/mol - 711.39 kJ/mol + 32.89 kJ/mol
ΔG° = -1467.3 kJ/mol
So, ΔG° for the reaction C2H6 (g) + O2 (g) → 2 CO2 (g) + 3 H2O (l) at 1 atm and 25 °C is -1467.3 kJ/mol.
Now, that's some calculating done with a touch of clowny humor!
To calculate ΔG° for the given reaction, we need to consider the stoichiometry of the reaction and use the formula:
ΔG° = Σ(nΔGf° products) - Σ(mΔGf° reactants)
where n and m are the stoichiometric coefficients and ΔGf° represents the standard Gibbs free energy of formation for each compound.
First, let's balance the equation:
C2H6 (g) + O2 (g) → CO2 (g) + H2O (l)
Balanced equation:
2 C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (l)
Now, we can calculate ΔG°:
ΔG° = [4(ΔGf° CO2) + 6(ΔGf° H2O)] - [2(ΔGf° C2H6) + 7(ΔGf° O2)]
Substituting the given values:
ΔG° = [4(-394.4 kJ/mol) + 6(-237.13 kJ/mol)] - [2(-32.89 kJ/mol) + 7(0 kJ/mol)]
Calculating:
ΔG° = (-1577.6 kJ/mol) + (-1422.78 kJ/mol) - (-65.78 kJ/mol)
ΔG° = -1577.6 kJ/mol - 1422.78 kJ/mol + 65.78 kJ/mol
ΔG° = -2934.6 kJ/mol
Therefore, the value of ΔG° for the reaction C2H6 (g) + O2 (g) → CO2 (g) + H2O (l) at 1 atm and 25 °C is -2934.6 kJ/mol.
To calculate the standard Gibbs free energy change (ΔG°) for the reaction, you need to balance the equation first. The balanced equation is:
2 C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (l)
Next, you can use the following formula to calculate ΔG°:
ΔG° = ΣnΔGf°(products) - ΣnΔGf°(reactants)
In this formula, Σn is the stoichiometric coefficient of each species and ΔGf° is the standard Gibbs free energy of formation for that species.
Now let's plug in the values you provided:
ΔG° = [4 * ΔGf°(CO2)] + [6 * ΔGf°(H2O)] - [2 * ΔGf°(C2H6) + 7 * ΔGf°(O2)]
ΔG° = [4 * (-394.4 kJ/mol)] + [6 * (-237.13 kJ/mol)] - [2 * (-32.89 kJ/mol) + 7 * 0 kJ/mol]
ΔG° = -1577.6 kJ/mol - 1422.78 kJ/mol + 65.78 kJ/mol
ΔG° = -1934.6 kJ/mol
Therefore, the ΔG° for the given reaction at 1 atm and 25 °C is -1934.6 kJ/mol.
You need to study Hess'law.
balance the equation first.
2C2H6 (g) + 7O2 (g) → 4CO2 (g) + 6H2O (l)
free energy=3(HfCO2)+6Hf(H20)-2HfC2H6)