The thermochemical equation for the combustion of hexane is shown below.
2 C6H14 (g) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (g) ΔH° = –8326 kJ
What is the enthalpy change for the combustion of 0.650 L C6H14 at STP?
8326 kJ is produced by 2*22.4L C6H14. So for 0.650 L we will produce
8326 kJ x (0.650L/2*22.4L) = ? kJ.
To find the enthalpy change for the combustion of 0.650 L of C6H14 at STP, we need to use the ideal gas law to calculate the number of moles of C6H14.
1. Calculate the number of moles of C6H14:
Since we know the volume and the conditions are stated as STP (standard temperature and pressure), we can use the ideal gas law:
PV = nRT
STP conditions:
P = 1 atm
V = 0.650 L
T = 273 K
R = 0.0821 L·atm/(K·mol)
Plugging in the values:
(1 atm) * (0.650 L) = n * (0.0821 L·atm/(K·mol)) * (273 K)
n = (1 atm * 0.650 L) / (0.0821 L·atm/(K·mol) * 273 K)
n ≈ 18.78 mol
So, the number of moles of C6H14 is approximately 18.78 mol.
2. Calculate the enthalpy change for the combustion:
Since the thermochemical equation is balanced, we know that for every 2 moles of C6H14 combusted, the enthalpy change is -8326 kJ.
So, for 18.78 mol of C6H14 combusted, the enthalpy change can be calculated using stoichiometry:
Enthalpy change = (-8326 kJ / 2 mol) * (18.78 mol / 2 mol)
Enthalpy change ≈ -39242.88 kJ
Therefore, the enthalpy change for the combustion of 0.650 L of C6H14 at STP is approximately -39242.88 kJ.
To find the enthalpy change for the combustion of 0.650 L of C6H14 at STP (Standard Temperature and Pressure), we need to use the given thermochemical equation and the ideal gas law to determine the amount of substance in moles.
First, let's convert the volume of C6H14 from liters to moles. To do this, we need to know the molar volume of an ideal gas at STP, which is approximately 22.4 L/mol:
0.650 L C6H14 × (1 mol/22.4 L) = 0.029 moles of C6H14
Now that we know the number of moles of C6H14, we can use the stoichiometry of the reaction to determine the enthalpy change. According to the balanced equation:
2 moles of C6H14 produce -8326 kJ of energy
Therefore, 0.029 moles of C6H14 would produce:
0.029 moles C6H14 × (-8326 kJ/2 moles C6H14) = -120.57 kJ
So, the enthalpy change for the combustion of 0.650 L of C6H14 at STP is approximately -120.57 kJ. Note that the negative sign indicates that the reaction is exothermic, meaning it releases heat energy.