A 1.000 g sample of ethanol, C2H5OH was burned in a coffee cup calorimeter containing 1220 mL of water. The temperature of the water in the calorimeter increased by 5.25C . What is the molar enthalpy of combustion in kJ/mol?
To calculate the molar enthalpy of combustion, we need to first find the amount of heat transferred to the water in the calorimeter.
The heat transferred can be determined using the equation:
q = mcΔT
Where:
q = heat transferred
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
The specific heat capacity of water is 4.18 J/g°C. Converting the mass of water from mL to g:
mass of water = 1220 mL = 1220 g
Now, we can calculate the heat transferred:
q = (1220 g)(4.18 J/g°C)(5.25°C)
q = 27,343 J
Next, we need to convert the units of heat to kJ:
q = 27,343 J ÷ 1000 = 27.343 kJ
Since the ethanol sample weighed 1.000 g, the molar enthalpy of combustion can be calculated using the equation:
Molar enthalpy of combustion = q ÷ n
Where:
q = heat transferred (27.343 kJ)
n = moles of ethanol
The molar mass of ethanol (C2H5OH) is:
C = 12.01 g/mol
H = 1.01 g/mol (x 5 atoms)
O = 16.00 g/mol
Molar mass of ethanol = 12.01 + (1.01 x 5) + 16.00 = 46.07 g/mol
Now, we can calculate the moles of ethanol:
moles of ethanol = 1.000 g ÷ 46.07 g/mol
moles of ethanol ≈ 0.0217 mol
Finally, we can calculate the molar enthalpy of combustion:
Molar enthalpy of combustion = 27.343 kJ ÷ 0.0217 mol
Molar enthalpy of combustion ≈ 1259.4 kJ/mol
Therefore, the molar enthalpy of combustion of ethanol is approximately 1259.4 kJ/mol.
To calculate the molar enthalpy of combustion, we need to use the equation:
ΔH = q / n
where ΔH is the molar enthalpy of combustion, q is the heat released, and n is the number of moles of ethanol.
First, let's calculate the heat released in the reaction using the equation:
q = m * C * ΔT
where q is the heat released, m is the mass of water, C is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
Mass of water (m) = 1220 g
Specific heat capacity of water (C) = 4.18 J/g°C
Change in temperature (ΔT) = 5.25°C
Substituting the given values into the equation, we get:
q = 1220 g * 4.18 J/g°C * 5.25°C
q ≈ 33126.15 J
Next, we need to calculate the number of moles of ethanol using its molecular weight.
The molecular weight of ethanol (C2H5OH) is:
2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.07 g/mol
Given:
Mass of ethanol = 1.000 g
Substituting the values into the equation, we get:
n = mass of ethanol / molecular weight
n = 1.000 g / 46.07 g/mol
n ≈ 0.0217 mol
Now we can substitute the values of q and n into the equation for molar enthalpy of combustion:
ΔH = q / n
ΔH = 33126.15 J / 0.0217 mol
ΔH ≈ 1523796.77 J/mol
Finally, converting the units from J/mol to kJ/mol:
ΔH ≈ 1523796.77 J/mol * (1 kJ / 1000 J)
ΔH ≈ 1523.8 kJ/mol
Therefore, the molar enthalpy of combustion of ethanol is approximately 1523.8 kJ/mol.
To find the molar enthalpy of combustion of ethanol, we need to calculate the heat transferred from the combustion to the water in the calorimeter.
First, we need to calculate the heat transferred using the formula:
q = m × c × ΔT
where:
q is the heat transferred,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
To calculate the mass of the water, we need to convert the volume of water from milliliters (mL) to grams (g). The density of water is approximately 1 g/mL. So, the mass of the water is:
mass = volume × density
mass = 1220 mL × 1 g/mL
mass = 1220 g
Next, we need to calculate the heat transferred:
q = 1220 g × 4.18 J/g°C × 5.25°C
q = 27182.5 J
Since the question asks for the molar enthalpy of combustion in kJ/mol, we need to convert the heat transferred from joules (J) to kilojoules (kJ):
q = 27182.5 J ÷ 1000
q = 27.18 kJ
Now, we need to find the number of moles of ethanol burned. The molar mass of ethanol (C2H5OH) is:
molar mass = (2 × atomic mass of carbon) + (6 × atomic mass of hydrogen) + atomic mass of oxygen
molar mass = (2 × 12.01 g/mol) + (6 × 1.01 g/mol) + 16.00 g/mol
molar mass = 46.07 g/mol
Since the mass of the ethanol sample burned is given as 1.000 g, the number of moles is:
moles = mass ÷ molar mass
moles = 1.000 g ÷ 46.07 g/mol
moles = 0.0217 mol
Finally, we can calculate the molar enthalpy of combustion:
molar enthalpy of combustion = heat transferred ÷ moles
molar enthalpy of combustion = 27.18 kJ ÷ 0.0217 mol
molar enthalpy of combustion = 1253.6 kJ/mol
Therefore, the molar enthalpy of combustion of ethanol is 1253.6 kJ/mol.