Question

Please, I need a step-by-step guidance on how to solve ths:

The temperature rises from 25°C to 29° in a bomb calorimeter when 3.50g of sucrose undergoes combustion in a bomb calorimteter. Calculate the heat of combustion of sucrose in kJ/mol sucrose. The heat capacity of the calorimeter is 4.90 kJ/°C. The molar mass of sucrose is 342.3g/mol.

I'm confused with the units.

Answer 1

\(q=mc \Delta T\\q=(3.5g)(4.90\ kJ/°C)(4°C)\\q=68.6\ kJ•g\)


then if I convert it to kj/mol
\(68.6\ kJ•g(1\ mol/342.3g)\) the unit that I will get is kj•mol not kj/mol

Answer 2

but if i will multiply it by the molar mass, i will get a unit of kj•g^2/mol

Answer 3

@zepdrix @pooja195

Answer 4

The set up looks like this:



The reaction gives off thermal energy which is absorbed by the water (not mentioned in the question) and the temperature rises.

The specific heat capcity of the calorimeter here includes the specific heat capacity of the water, so the formula for this is reduced to:

\(\sf \large q_{absorbed}=C_{calorimeter}*\Delta T\)

The mass of the sugar and it's heat of combustion is used to find exactly how much thermal energy was given to the water. The formula for this is:

\(\sf \large q_{released}=-moles*\Delta H_{comb}\)

Solving for \(\sf \Delta H_{comb}\) gives: \(\sf \large \Delta H=-\dfrac{q_{released}}{moles }\)

(NOTE: sucrose needs to be in moles not grams look at the units of the heat of combustion)

The energy absorbed by the water and energy given off are related by:

\(\sf \large q_{absorbed}=-q_{released}\)

So the combined formulas are:

lving for \(\sf \Delta H_{comb}\) gives: \(\sf \large \Delta H=-\dfrac{q_{released}}{moles }=\dfrac{C_{calorimeter}*\Delta T}{moles_{sucrose}}\)