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: ΔH°f[CO2(g)] = –393.5 kJ/mol, ΔH°f[H2O(l)] = –285.8 kJ/mol, ΔH°f[H2O(g)] = –241.8 kJ/mol)
To find the standard enthalpy of formation (ΔH°f) of styrene at 25°C, we can use the standard enthalpies of formation of the products and reactants involved in the combustion reaction.
The balanced chemical equation for the combustion of styrene is:
C8H8 + 12O2 → 8CO2 + 4H2O
From the given information, we have:
ΔH°f[CO2(g)] = –393.5 kJ/mol
ΔH°f[H2O(l)] = –285.8 kJ/mol
We need to account for the enthalpy change for water in its liquid form, but we are given the enthalpy change for water in its gas form (ΔH°f[H2O(g)] = –241.8 kJ/mol).
We can calculate the enthalpy change for water in its liquid form by subtracting the enthalpy change for water in its gaseous form from the enthalpy change for water in its liquid form:
ΔH°f[H2O(l)] = ΔH°f[H2O(g)] - ΔH°vap[H2O]
The enthalpy of vaporization (ΔH°vap[H2O]) for water at 25°C is 40.0 kJ/mol. Therefore:
ΔH°f[H2O(l)] = –241.8 kJ/mol - 40.0 kJ/mol = –281.8 kJ/mol
Now, we can calculate the ΔH°f of styrene at 25°C. The enthalpy change for the combustion of styrene is given as 42.62 kJ per gram of styrene.
First, we need to convert grams of styrene to moles using the molar mass of styrene.
The molar mass of styrene (C8H8) is 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) = 104.16 g/mol
Next, we can calculate the ΔH°f of styrene using the following formula:
ΔH°f = ΔH°products - ΔH°reactants
ΔH°products = (8 * ΔH°f[CO2(g)]) + (4 * ΔH°f[H2O(l)])
ΔH°reactants = ΔH°f[styrene]
ΔH°f = (8 * –393.5 kJ/mol) + (4 * –281.8 kJ/mol) – 42.62 kJ/mol
Calculating the above expression gives us the result:
ΔH°f = –3171.68 kJ/mol
Therefore, the standard enthalpy of formation of styrene at 25°C is –3171.68 kJ/mol.
To find the standard enthalpy of formation of styrene, we need to use the given enthalpy values and the balanced combustion equation for styrene.
The balanced combustion equation for styrene is as follows:
C8H8(l) + 9O2(g) → 8CO2(g) + 4H2O(l)
First, we need to calculate the enthalpy change for the formation of 8 moles of CO2 and 4 moles of H2O, which can be calculated using the enthalpy of formation values given:
ΔH1 = 8 * ΔH°f[CO2(g)] + 4 * ΔH°f[H2O(l)]
ΔH1 = 8 * (-393.5 kJ/mol) + 4 * (-285.8 kJ/mol)
ΔH1 = -3148 kJ
Next, we need to calculate the enthalpy change for the combustion of 1 mole of styrene:
ΔH2 = 42.62 kJ/g * (molar mass of styrene)
ΔH2 = 42.62 kJ/g * (104.15 g/mol) [molar mass of styrene]
ΔH2 = -4444 kJ/mol
Finally, we can find the standard enthalpy of formation of styrene using the equation:
ΔH°f[C8H8(l)] = ΔH2 - ΔH1
ΔH°f[C8H8(l)] = -4444 kJ/mol - (-3148 kJ/mol)
ΔH°f[C8H8(l)] = -1296 kJ/mol
So, the standard enthalpy of formation of styrene at 25°C is -1296 kJ/mol.
C8H8 + 10 O2 ==> 8CO2 + 4H2O
42.62 kJ/gram. Convert to kJ/mole which will be delta Hrxn. Then
DHrxn = (n*DHproducts)-(DHreactants) and solve for DH styrene/mol.