What would you do in order to solve this question: "The enthalpy of combustion for benzoic acid is -3223.6kj/mol. Use the values of the standard enthalpies of formation of liquid water and CO2 to calculate standard enthalpy of formation of benzoic acid."
To calculate the standard enthalpy of formation of benzoic acid (ΔHf°), you can use the values of the standard enthalpies of formation (ΔHf°) of liquid water and carbon dioxide (CO2). The equation for the combustion of benzoic acid is:
C6H5COOH + O2 -> CO2 + H2O
To solve this question, follow these steps:
Step 1: Determine the balanced equation
Balanced combustion equation for benzoic acid is:
C6H5COOH + 15/2 O2 -> 7 CO2 + 3 H2O
Step 2: Calculate the enthalpy change of the reaction
To calculate the enthalpy change (ΔH) for this reaction, you need to know the enthalpies of formation of the reactants and products. The enthalpy change is given by:
ΔH = ΣnΔHf°(products) - ΣnΔHf°(reactants)
Where ΔHf° is the standard enthalpy of formation and n is the stoichiometric coefficient.
Step 3: Calculate the enthalpy change of the combustion reaction
ΔH = (7ΔHf°(CO2) + 3ΔHf°(H2O)) - (ΔHf°(C6H5COOH) + 15/2ΔHf°(O2))
Step 4: Substitute the given values
The enthalpy of combustion for benzoic acid is given as -3223.6 kJ/mol.
-3223.6 kJ/mol = (7ΔHf°(CO2) + 3ΔHf°(H2O)) - (ΔHf°(C6H5COOH) + 15/2ΔHf°(O2))
Step 5: Solve
Rearrange the equation and isolate ΔHf°(C6H5COOH):
ΔHf°(C6H5COOH) = 7ΔHf°(CO2) + 3ΔHf°(H2O) + 15/2ΔHf°(O2) + 3223.6 kJ/mol
Step 6: Substitute known values
Substitute the known values of the standard enthalpies of formation. The standard enthalpies of formation of carbon dioxide (CO2) and liquid water (H2O) are -393.5 kJ/mol and -285.8 kJ/mol, respectively. The standard enthalpy of formation of oxygen (O2) is 0 kJ/mol.
ΔHf°(C6H5COOH) = 7(-393.5 kJ/mol) + 3(-285.8 kJ/mol) + 15/2(0 kJ/mol) + 3223.6 kJ/mol
Step 7: Calculate
Calculate the value of ΔHf°(C6H5COOH) using the equation from step 6.
ΔHf°(C6H5COOH) = -393.5 kJ/mol + -285.8 kJ/mol + 3223.6 kJ/mol
ΔHf°(C6H5COOH) = 1107.7 kJ/mol
Therefore, the standard enthalpy of formation of benzoic acid is 1107.7 kJ/mol.
To calculate the standard enthalpy of formation of benzoic acid using the given enthalpy of combustion and standard enthalpies of formation, you need to apply the Hess's law and use the following equation:
ΔH°f (benzoic acid) = Σ(nΔH°f [products]) - Σ(mΔH°f [reactants])
1. Determine the balanced equation for the combustion of benzoic acid.
C6H5COOH + nO2 -> mCO2 + pH2O
2. Assign coefficients (n, m, p) to balance the equation.
3. Find the molar enthalpies of formation for the products (CO2 and H2O) and the reactant (benzoic acid).
The molar enthalpies of formation for CO2 and H2O are known values. You can find them in a reliable source or reference book. Let's assume:
ΔH°f (CO2) = -393.5 kJ/mol
ΔH°f (H2O(l)) = -285.8 kJ/mol
4. Substitute the values into the equation and solve for the standard enthalpy of formation of benzoic acid.
ΔH°f (benzoic acid) = (mΔH°f [CO2]) + (pΔH°f [H2O(l)]) - (nΔH°f [benzoic acid])
ΔH°f (benzoic acid) = (m * -393.5 kJ/mol) + (p * -285.8 kJ/mol) - (n * -3223.6 kJ/mol)
Make sure to substitute the correct coefficients (m, n, p) for the balanced equation. Finally, calculate the value and include the proper units to obtain the standard enthalpy of formation of benzoic acid.
The combustion of benzoic acid is
C7H6O2 + 15/2 O2 ==> 7CO2 + 3H2O and since you want the enthalpy of FORMATION (and not combustion), reverse the equation (and the sign for dH combustion) to obtain
7CO2 + 3H2O ==> C7H6O2 dH = +3223.6 kJ
Then dHrxn = (n*dHf benzoic acid)- (n*dHf products). Substitute 3223.6 kJ for dHrxn and substitute values for dHf H2O and dHf CO2 and calculate dHf benzoic acid.