Styrene, C8H8, is one of the substances used in the production of synthetic rubber. When styrene burns in oxygen to form carbon dioxide and liquid water under standard-state conditions at 25°C, 42.62 kJ are released per gram of styrene. Find the standard enthalpy of formation of styrene at 25°C.
(Given: [CO2(g)] = −393.5 kJ/mol, [H2O(l)] = −285.8 kJ/mol, [H2O(g)] = −241.8 kJ/mol)
Thank you so much!!!!
To find the standard enthalpy of formation of styrene at 25°C, we can use the equation:
ΔH = ΣnΔHf(products) - ΣmΔHf(reactants)
Where:
ΔH is the standard enthalpy change for the reaction,
ΔHf is the standard enthalpy of formation,
n and m are the stoichiometric coefficients of the products and reactants.
The balanced chemical equation for the combustion of styrene is:
C8H8 + 12 O2 -> 8 CO2 + 4 H2O
Using the given values for the enthalpies of formation:
[CO2(g)] = −393.5 kJ/mol
[H2O(l)] = −285.8 kJ/mol
We can calculate the standard enthalpy change of the reaction as follows:
ΔH = (8 * [CO2(g)]) + (4 * [H2O(l)]) - [C8H8]
ΔH = (8 * -393.5 kJ/mol) + (4 * -285.8 kJ/mol) - 42.62 kJ/g
ΔH = -3148 kJ/mol + (-1143.2 kJ/mol) - 42.62 kJ/g
ΔH = -4333.8 kJ/mol - 42.62 kJ/g
Now, we need to convert the units from kJ/g to kJ/mol for styrene.
The molar mass of styrene (C8H8) can be calculated as follows:
(8 * atomic mass of carbon) + (8 * atomic mass of hydrogen)
Molar mass of carbon = 12.01 g/mol
Molar mass of hydrogen = 1.008 g/mol
Molar mass of styrene = (8 * 12.01 g/mol) + (8 * 1.008 g/mol)
= 104.08 g/mol
To convert kJ/g to kJ/mol, we divide by the molar mass of styrene:
ΔH = (-4333.8 kJ/mol - 42.62 kJ/g) / 104.08 g/mol
Now we can calculate the standard enthalpy of formation of styrene at 25°C.
To find the standard enthalpy of formation of styrene at 25°C, we can use the Hess's Law of constant heat summation. This law states that the change in enthalpy for a chemical reaction is independent of the pathway between the initial and final states.
Since we are given the enthalpies of formation for carbon dioxide and liquid water, we can use them to calculate the enthalpy change for the combustion of styrene. The balanced equation for the combustion of styrene is:
C8H8 + 12O2 -> 8CO2 + 4H2O
We can write the enthalpy change for this reaction as follows:
ΔH(styrene combustion) = (8 * ΔH(CO2)) + (4 * ΔH(H2O))
Substituting the given values for enthalpies of formation:
ΔH(styrene combustion) = (8 * -393.5 kJ/mol) + (4 * -285.8 kJ/mol)
Calculating:
ΔH(styrene combustion) = -3148.0 kJ/mol + -1143.2 kJ/mol
= -4291.2 kJ/mol
We know that 42.62 kJ of heat is released per gram of styrene, so we need to calculate the molar mass of styrene to find the standard enthalpy of formation per mole of styrene.
The molar mass of styrene (C8H8) can be calculated as follows:
Molar mass of carbon (C) = 12.01 g/mol
Molar mass of hydrogen (H) = 1.01 g/mol
Molar mass of styrene = (8 * 12.01 g/mol) + (8 * 1.01 g/mol)
= 104.08 g/mol
Now, we can calculate the standard enthalpy of formation of styrene (ΔHf) per mole of styrene by dividing the enthalpy change for the combustion by the molar mass of styrene:
ΔHf = ΔH(styrene combustion) / Molar mass of styrene
= -4291.2 kJ/mol / 104.08 g/mol
Calculating:
ΔHf = -41.21 kJ/g
Therefore, the standard enthalpy of formation of styrene at 25°C is -41.21 kJ/g.
C8H8 + 10 O2(g) ==> 8CO2(g) + 4H2O(l)
delta Hrxn = (8*deltaHf CO2 + 4*deltaHf H2O) - (1*deltaHfC8H8)
You can look up CO2 and H2O in tables. deltaHf for oxygen will be zero since that is its standard state and at 25 C.I would convert delta H rxn from kJ/g to kJ/mol since all of the numbers you look up for CO2 and water are in kJ/mol. The answer for delta Hf for styrene will be in kJ/mol.