At constant volume, the heat of combustion of a particular compound is –3209.0 kJ/mol. When 1.829 g of this compound (molar mass = 159.29 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 6.753 °C. What is the heat capacity (calorimeter constant) of the calorimeter?
How much heat was evolved in the combustion? That's 3209.0 kJ/mol x (1.829/159.29) = q = ?
Then q = Ccal*delta T.
Solve for Ccal.
first you do n=m/M to get your moles then use you moles and the -3209.0 to geyt your kj then solve for the heat capacity by using the equation q=mct to rearrange to c=q/mt
To find the heat capacity of the calorimeter, we can use the equation:
q = C * ΔT
where:
q = heat transferred to the calorimeter
C = heat capacity of the calorimeter
ΔT = change in temperature
In this case, we know the heat transferred (q) and the change in temperature (ΔT), so we can rearrange the equation to solve for the heat capacity (C):
C = q / ΔT
First, we need to convert the heat transferred from kJ to J and the mass from grams to moles.
Step 1: Convert heat transferred from kJ to J
Heat transferred (q) = -3209.0 kJ/mol
Since there are 1000 J in 1 kJ, we can convert:
q = -3209.0 kJ/mol * 1000 J/kJ = -3,209,000 J/mol
Step 2: Convert the mass from grams to moles
Mass of the compound = 1.829 g
Molar mass of the compound = 159.29 g/mol
We use the formula:
moles = mass (g) / molar mass (g/mol)
moles = 1.829 g / 159.29 g/mol = 0.011485 mol
Step 3: Calculate the heat capacity of the calorimeter
Now, we can substitute the values into the equation:
C = q / ΔT
C = (-3,209,000 J/mol) / 6.753 °C
Note: ΔT should be converted to Kelvin since we are dealing with temperature differences.
ΔT = 6.753 °C + 273.15 K = 280.903 K
C = -3,209,000 J/mol / 280.903 K ≈ -11,428.3 J/(mol·K)
Therefore, the heat capacity (calorimeter constant) of the calorimeter is approximately -11,428.3 J/(mol·K).