Question

At constant volume, the heat of combustion of a particular compound, compound A, is –3568.0 kJ/mol. When 1.411 g of compound A (molar mass = 115.27 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.379 °
c. Using this data, what is the heat capacity (calorimeter constant) of the calorimeter?

Answer 1

–3568.0 kJ/mol * 1.411 g / 115.27 g/mol = -43.67 kJ

Heat capacity of the calorimeter: -43.67 kJ / 3.379 ° = -12.92 kJ/deg C

Not 100% sure about my answer

Answer 2

Answer : The heat capacity (calorimeter constant) of the calorimeter is, $12.92kJ/^oC$

Explanation:

First we have to calculate the moles of compound A.

$\text{Moles of compound A}=\frac{\text{Mass of compound A}}{\text{Molar mass of compound A}}$

$\text{Moles of compound A}=\frac{1.411g}{115.27g/mol}=0.01224mol$

Now we have to calculate the heat of combustion of compound A for 0.01224 mol.

As, 1 mole of compound A has heat of combustion = 3568.0 kJ

So, 0.01224 mole of compound A has heat of combustion = 0.01224 × 3568.0 kJ

                                                                                                 = 43.67 kJ

Now we have to calculate the heat capacity (calorimeter constant) of the calorimeter.

$q=c\times \Delta T$

where,

q = heat of combustion = 43.67 kJ

c = heat capacity = ?

$\Delta T$ = change in temperature = $3.379^oC$

Now put all the given values in the above expression, we get:

$43.67kJ=c\times (3.379^oC)$

$c=12.92kJ/^oC$

Therefore, the heat capacity (calorimeter constant) of the calorimeter is, $12.92kJ/^oC$