The combustion reaction for benzoic acid
C6H5CO2H(s) +
15
2
O2(g) →
7 CO2(g) + 3 H2O(ℓ)
has ∆H0
= −3226.7 kJ/mole. Use Hess’s
Law to calculate ∆H0
f
for benzoic acid.
To calculate ∆H0f for benzoic acid, we need to use the given ∆H0 value for the combustion reaction and the ∆H0f values for carbon dioxide (CO2) and water (H2O). The key idea behind Hess's Law is that the enthalpy change for a reaction can be calculated by combining the enthalpy changes of other reactions.
The balanced equation for the combustion of benzoic acid is:
C6H5CO2H(s) + 15/2 O2(g) → 7 CO2(g) + 3 H2O(ℓ)
To use Hess's Law, we need to manipulate this equation so that it matches the known reactions. Let's start by reversing the reaction for the formation of CO2:
CO2(g) → C(s) + O2(g)
This reaction has an enthalpy change of ∆H0f for carbon dioxide (CO2) = -393.5 kJ/mol.
Next, let's reverse the reaction for the formation of water:
H2O(ℓ) → H2(g) + 1/2 O2(g)
This reaction has an enthalpy change of ∆H0f for water (H2O) = -285.8 kJ/mol.
Now, we will add these manipulated reactions to the given combustion reaction equation, noting that we need to multiply the enthalpy changes for the manipulated reactions by the appropriate coefficients:
C6H5CO2H(s) + 15/2 O2(g) → 7 CO2(g) + 3 H2O(ℓ)
First manipulated reaction:
7 CO2(g) + 7 C(s) + 7 O2(g)
Second manipulated reaction:
3 H2O(ℓ) + 3 H2(g) + 3/2 O2(g)
Combining these reactions, we get:
C6H5CO2H(s) + 7 C(s) + 7 H2(g) + 11/2 O2(g) → 7 CO2(g) + 3 H2O(ℓ)
Now, we can write the enthalpy change equation using the enthalpy values:
∆H0 = ∆H0f for benzoic acid + 7*∆H0f for carbon + 7*∆H0f for hydrogen + 11/2*∆H0 for oxygen - 7*∆H0f for CO2 - 3*∆H0f for water
Plugging in the values:
-3226.7 kJ/mol = ∆H0f for benzoic acid + 7*(-393.5 kJ/mol) + 7*0 kJ/mol + 11/2*0 kJ/mol - 7*0 kJ/mol - 3*(-285.8 kJ/mol)
Simplifying the equation:
-3226.7 kJ/mol = ∆H0f for benzoic acid - 2754.5 kJ/mol
∆H0f for benzoic acid = -3226.7 kJ/mol + 2754.5 kJ/mol
∆H0f for benzoic acid = -472.2 kJ/mol
Therefore, the enthalpy change of formation (∆H0f) for benzoic acid is -472.2 kJ/mol.
To calculate ΔH°f for benzoic acid using Hess's Law, we need to use the known enthalpies of formation for the other substances involved in the combustion reaction.
Hess's Law states that the enthalpy change of a reaction is the same whether it occurs in one step or multiple steps as long as the initial and final conditions are the same. In this case, we'll use the fact that the enthalpy change in the combustion reaction equals the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.
The enthalpy of formation (∆H°f) is the change in enthalpy when one mole of a compound is formed under standard conditions (25°C and 1 atm pressure).
To calculate ∆H°f for benzoic acid, we'll break down the combustion reaction into two steps:
Step 1: Formation of carbon dioxide and water:
C6H5CO2H(s) + 15/2 O2(g) → 7 CO2(g) + 3 H2O(ℓ)
Step 2: Formation of benzoic acid:
C(s) + H2(g) + CO2(g) → C6H5CO2H(s)
Now, we need to find the enthalpies of formation for each substance involved in these steps.
1. The enthalpies of formation for carbon dioxide (CO2) and water (H2O) are tabulated values:
∆H°f(CO2) = -393.5 kJ/mol
∆H°f(H2O) = -285.8 kJ/mol
2. The enthalpy of formation for carbon (C) can be assumed to be zero since it is in its standard state.
3. The enthalpy of formation for hydrogen gas (H2) is also a tabulated value:
∆H°f(H2) = 0 kJ/mol
Now, we can calculate the enthalpy change for each step:
Step 1:
7 mol CO2 (∆H°f(CO2)) + 3 mol H2O (∆H°f(H2O))
Step 2:
1 mol C (∆H°f(C)) + 1 mol H2 (∆H°f(H2)) + 1 mol CO2 (∆H°f(CO2))
Finally, we can calculate ∆H°f for benzoic acid:
∆H°f(benzoic acid) = ∆H°f(C6H5CO2H) = ∆H°f(C6H5CO2H) - [7 mol CO2 (∆H°f(CO2)) + 3 mol H2O (∆H°f(H2O)) - 1 mol C (∆H°f(C)) - 1 mol H2 (∆H°f(H2)) - 1 mol CO2 (∆H°f(CO2))]
Plugging in the values:
∆H°f(benzoic acid) = -3226.7 kJ/mol - [7 (-393.5 kJ/mol) + 3 (-285.8 kJ/mol) - 0 - 0 - (-393.5 kJ/mol)]
Simplifying the equation gives:
∆H°f(benzoic acid) = -3226.7 kJ/mol + [2754.5 kJ/mol + 857.4 kJ/mol + 0 + 0 + 393.5 kJ/mol]
Calculating the final value will give the enthalpy of formation for benzoic acid (∆H°f):
∆H°f(benzoic acid) = -3226.7 kJ/mol + 4005.4 kJ/mol
∆H°f(benzoic acid) = 778.7 kJ/mol
Therefore, ∆H°f for benzoic acid is 778.7 kJ/mol.