Propane (C3H8) burns according to the following balanced equation:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
Calculate ΔH∘rxn for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of gaseous propane is -103.9 kJ/mol.)
Now you need dHof fpr H2O gas.
dHrxn = (n*dH products) - (n*dH reactants)
To calculate ΔH∘rxn for the given reaction, we need to use the standard enthalpies of formation of the reactants and products.
Given:
Standard enthalpy of formation of gaseous propane (C3H8): -103.9 kJ/mol
The balanced equation for the reaction is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
We can calculate the ΔH∘rxn using the equation:
ΔH∘rxn = ΣnΔH∘f(products) - ΣnΔH∘f(reactants)
Let's calculate it step-by-step:
Step 1: Calculate the ΔH∘f for the reactants.
ΔH∘f(C3H8) = -103.9 kJ/mol
Step 2: Calculate the ΔH∘f for the products.
ΔH∘f(CO2) = 0 kJ/mol (Standard enthalpy of formation for CO2 is often taken as zero because it is the most stable form of carbon)
ΔH∘f(H2O) = -241.8 kJ/mol (Standard enthalpy of formation for H2O)
Step 3: Calculate the moles of reactants and products.
Reactants:
1 mole of C3H8
5 moles of O2
Products:
3 moles of CO2
4 moles of H2O
Step 4: Calculate the ΔH∘rxn using the formula.
ΔH∘rxn = (3 * ΔH∘f(CO2) + 4 * ΔH∘f(H2O)) - (ΔH∘f(C3H8) + 5 * ΔH∘f(O2))
= (3 * 0 kJ/mol + 4 * -241.8 kJ/mol) - (-103.9 kJ/mol + 5 * 0 kJ/mol)
= (-967.2 kJ/mol) - (-103.9 kJ/mol)
= -863.3 kJ/mol
Therefore, the ΔH∘rxn for the given reaction is -863.3 kJ/mol.
To calculate ΔH∘rxn for the reaction using standard enthalpies of formation, you need to know the standard enthalpies of formation for all the reactants and products involved.
Given:
Standard enthalpy of formation of gaseous propane (C3H8): -103.9 kJ/mol
The standard enthalpy of formation for a substance represents the change in enthalpy when one mole of the substance is formed from its elements in their standard states at a specified temperature and pressure.
Using the balanced equation, C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g), the enthalpy change ΔH∘rxn can be calculated by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products.
Step 1: Write down the balanced equation:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Step 2: Determine the enthalpy change for each compound using the enthalpies of formation:
ΔH∘rxn = (3 × ΔH∘f[CO2]) + (4 × ΔH∘f[H2O]) - (ΔH∘f[C3H8] + 5 × ΔH∘f[O2])
Step 3: Substitute the given values:
ΔH∘f[CO2] = 0 kJ/mol (because it is the standard state)
ΔH∘f[H2O] = 0 kJ/mol (because it is the standard state)
ΔH∘f[C3H8] = -103.9 kJ/mol (given)
Step 4: Calculate ΔH∘rxn:
ΔH∘rxn = (3 × 0) + (4 × 0) - (-103.9) - (5 × 0)
Therefore, ΔH∘rxn = -103.9 kJ/mol
So, the standard enthalpy change for the combustion of propane is -103.9 kJ/mol.