To find the standard enthalpy of formation (ΔHf) of butane, we can use the given thermochemical equation for its combustion:
2C4H10 + 13O2 -> 8CO2 + 10H2O ΔH = -2880 kJ/mol
We can start by finding the ΔHf of butane. To do this, we have to use the enthalpies of formation of carbon dioxide (CO2) and water (H2O).
The enthalpy change for the reaction can be given by:
ΔH = Σ(ΔHf of products) - Σ(ΔHf of reactants)
-2880 kJ/mol = (8 mol x ΔHf of CO2) + (10 mol x ΔHf of H2O) - (2 mol x ΔHf of C4H10)
Using the given ΔHf values:
-2880 kJ/mol = (8 mol x (-394 kJ/mol)) + (10 mol x (-285.8 kJ/mol)) - (2 mol x ΔHf of C4H10)
Now let's solve for ΔHf of C4H10:
-2880 kJ/mol = (-3152 kJ) + (-2858 kJ) - (2 mol x ΔHf of C4H10)
-2880 kJ/mol = -6008 kJ - (2 mol x ΔHf of C4H10)
-2880 kJ/mol + 6008 kJ = -2 mol x ΔHf of C4H10
3128 kJ = 2 mol x ΔHf of C4H10
ΔHf of C4H10 = 1564 kJ/mol
So, the standard enthalpy of formation of butane (C4H10) is 1564 kJ/mol.
Now, let's calculate the mass of butane needed to raise the temperature of 100.0 g of H2O by 25.0 degrees C.
First, we can calculate the energy required to heat the water using the formula:
q = mcΔT
where:
q = heat absorbed (in joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (in °C)
Plugging in the values:
q = (100.0 g) x (4.18 J/g°C) x (25.0 °C)
q = 10450 J
Now, we can convert Joules to kilojoules because the enthalpy of formation is given in kJ:
10450 J ÷ 1000 = 10.45 kJ
Finally, we can calculate the mass of butane needed using the enthalpy change from the combustion of butane:
10.45 kJ ÷ 1564 kJ/mol = 0.0067 moles
Now, we can convert moles to grams of butane using the molar mass of butane:
1 mole of C4H10 = 58.12 g
0.0067 moles x 58.12 g/mol = 0.387 g
Therefore, approximately 0.387 grams of butane is needed to raise the temperature of 100.0 g of H2O by 25.0 degrees C.
