Sucrose, C12H22O11, is table sugar, and it has a ∆Hc of –5639.7 kJ/mol. Determine the ∆Hf of sucrose given
∆Hf (H2O(l)) = –285.8 kJ/mol, and ∆Hf (CO2(g)) = –393.5 kJ/mol .
C12H22O11(s) + O2(g) → CO2(g) + H2O(l)
I don't know your definition of delta Hc but I don't know that it is needed to solve for dHf sucrose.
C12H22O11(s) + (35/2)O2(g) →12CO2(g) + 11H2O(l)
dHrxn = (n*dHf products) - (n*dHf reactants)
Plug in the values and solve. Post your work if you get stuck.
Well, it looks like we're trying to determine the enthalpy of formation (∆Hf) of sucrose. To do that, we'll need to use the enthalpies of formation of the other substances involved in the reaction.
So, first let's write down the balanced equation again:
C12H22O11(s) + O2(g) → CO2(g) + H2O(l)
Now, we can look at the substances involved:
For CO2(g), ∆Hf = -393.5 kJ/mol.
For H2O(l), ∆Hf = -285.8 kJ/mol.
And for C12H22O11(s), we're trying to find ∆Hf.
Since we're not given the enthalpy of formation of O2(g), we'll assume it's zero. Now, let's set up the equation:
∆Hf (C12H22O11) + ∆Hf (O2) = ∆Hf (CO2) + ∆Hf (H2O)
Substituting the values we know:
∆Hf (C12H22O11) + 0 = -393.5 kJ/mol + (-285.8 kJ/mol)
Simplifying:
∆Hf (C12H22O11) = -679.3 kJ/mol
So, the ∆Hf of sucrose is approximately -679.3 kJ/mol. Keep in mind that this is a rough estimate and can vary depending on experimental conditions.
And remember, when life gives you sugar, make a delicious dessert and share it with your friends!
To determine the ∆Hf of sucrose, we can use the given equation:
C12H22O11(s) + O2(g) → CO2(g) + H2O(l)
The equation represents the combustion of sucrose, which forms carbon dioxide (CO2) and water (H2O). We need to use the standard enthalpy of formation (∆Hf) values of the products and reactants to calculate ∆Hf of sucrose.
First, let's calculate the ∆Hf for the reactant, C12H22O11(s):
∆Hf(C12H22O11) = Σ∆Hf(products) - Σ∆Hf(reactants)
Given:
∆Hf(CO2) = -393.5 kJ/mol
∆Hf(H2O) = -285.8 kJ/mol
Plugging in the values, we get:
∆Hf(C12H22O11) = ∆Hf(CO2) + ∆Hf(H2O) - ∆Hf(O2)
Since O2(g) is the naturally occurring form of oxygen, its ∆Hf is zero.
∆Hf(C12H22O11) = -393.5 kJ/mol + (-285.8 kJ/mol) - 0 kJ/mol
∆Hf(C12H22O11) = -393.5 kJ/mol - 285.8 kJ/mol
∆Hf(C12H22O11) = -679.3 kJ/mol
Therefore, the ∆Hf of sucrose (C12H22O11) is -679.3 kJ/mol.
To determine the enthalpy of formation (∆Hf) of sucrose (C12H22O11), we can use the Hess's law and the given enthalpies of combustion (∆Hc) of C12H22O11, enthalpy of formation of water (∆Hf(H2O(l))), and enthalpy of formation of carbon dioxide (∆Hf(CO2(g))).
The equation for the combustion of sucrose is as follows:
C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)
To find the enthalpy of formation of sucrose, we need to reverse the given equation to obtain the formation of sucrose from carbon dioxide and water. We also need to multiply the enthalpies of formation of CO2 and H2O by their stoichiometric coefficients in the balanced equation.
Reversed equation:
12CO2(g) + 11H2O(l) → C12H22O11(s) + 12O2(g)
Using Hess's law, the enthalpy change of the reaction can be calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products:
∆Hf(C12H22O11) = [12 × ∆Hf(CO2)] + [11 × ∆Hf(H2O)] - [12 × ∆Hc(C12H22O11)]
Substituting the given values:
∆Hf(C12H22O11) = [12 × (-393.5 kJ/mol)] + [11 × (-285.8 kJ/mol)] - [12 × (-5639.7 kJ/mol)]
Calculating the equation:
∆Hf(C12H22O11) = -4722 kJ/mol + (-3144 kJ/mol) - (-67676.4 kJ/mol)
∆Hf(C12H22O11) = -4722 kJ/mol - 3144 kJ/mol + 67676.4 kJ/mol
∆Hf(C12H22O11) = 59810.4 kJ/mol
Therefore, the enthalpy of formation (∆Hf) of sucrose (C12H22O11) is 59810.4 kJ/mol.