The enthalpy of combustion of benzoic acid (C6H5CO2H) is -3228 kJ/mol. The burning of 1.698 g of benzoic acid in a calorimeter causes the temperature to increase by 2.865°C. What is the heat capacity (in kJ/°C) of the calorimeter?
To find the heat capacity of the calorimeter, we can use the formula:
q = mcΔT
where q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
First, we need to find the heat transferred (q) when 1.698 g of benzoic acid is burned. We can calculate this using the enthalpy of combustion.
Enthalpy change (ΔH) of combustion = -3228 kJ/mol
First, let's find the number of moles of benzoic acid burned:
Molar mass of benzoic acid (C6H5CO2H):
C: 12.01 g/mol
H: 1.01 g/mol
O: 16.00 g/mol
Atomic mass of benzoic acid = (12.01 * 7) + (1.01 * 6) + (16.00 * 2) = 122.12 g/mol
Number of moles = Mass / Molar mass = 1.698 g / 122.12 g/mol
Now, we can find the heat transferred:
q = ΔH * moles
= -3228 kJ/mol * (1.698 g / 122.12 g/mol)
Next, we can calculate the heat capacity of the calorimeter using the formula q = mcΔT:
q = mcΔT
Rearranging the equation to solve for c:
c = q / (m * ΔT)
The mass (m) of the calorimeter is typically given or can be determined by weighing the calorimeter.
Now, substitute the known values into the equation to find the heat capacity (c) of the calorimeter.
To find the heat capacity of the calorimeter, we can use the equation:
Enthalpy Change = Heat Capacity * Temperature Change
First, we need to calculate the enthalpy change (ΔH) for the combustion of 1.698 g of benzoic acid.
To convert grams to moles, we need to find the molar mass of benzoic acid (C6H5CO2H):
C = 6 * 12.01 g/mol = 72.06 g/mol
H = 5 * 1.01 g/mol = 5.05 g/mol
O = 2 * 16.00 g/mol = 32.00 g/mol
Total molar mass of benzoic acid = 72.06 + 5.05 + 32.00 = 109.11 g/mol
Now we can calculate the number of moles of benzoic acid:
moles = mass / molar mass = 1.698 g / 109.11 g/mol = 0.0156 mol
Next, we can calculate the enthalpy change (ΔH) for the combustion of 0.0156 mol of benzoic acid:
ΔH = -3228 kJ/mol * 0.0156 mol = -50.34 kJ
Now we have both the enthalpy change (ΔH) and the temperature change (ΔT), we can rearrange the equation to solve for the heat capacity (C):
C = ΔH / ΔT = -50.34 kJ / 2.865°C = -17.58 kJ/°C
However, the negative sign indicates that the heat capacity of the calorimeter is negative. This means that the calorimeter loses heat as the temperature increases, which is unusual. It is possible that there are some sources of heat loss in the experiment that are not properly accounted for.