Chemistry Calorimeter/ heat capacity

Question

The combustion of toluene has delta E rxn of -3.91x10^3 kj/mol. When 1.55g of toluene (C7H8) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 Celsius to 37.57 Celsius. Find the heat capacity of the bomb calorimeter.

Formula for heat capacity: C = q/delta T ( change in temperature)

PS: we have change the in temperature but I'm not sure what the heat added would be, that is what is confusing me. Thank you.

Answer 1

molar mass toluene = 92.14 g / mol

moles toluene present = 1.55 g x 1 mol / 92.14 g = 0.01682 mols

heat (q) generated from 0.01682 mols toluene = 0.01682 mols x 3.91x10³ kJ/mol = 65.77 kJ

C = q / ∆T

C = ?

q = 65.77 kJ

∆T = change in temperature = 37.57 - 23.12 = 14.45º

C = 65.77 kJ / 14.45º

C = 4.56 kJ / degree

Answer 2

Great job identifying the correct equation for heat capacity!

We are given the E_rxn = -3.91 x 10³ kJ/mol. This number is negative, meaning the reaction is giving off energy. This makes sense because we know the temperature of the calorimeter is going up. The heat added to the calorimeter is equal to the heat given off by the exothermic reaction.

We need to find q to use in our heat capacity equation. q has units of Joules.

We can convert E_rxn to q using the mass of toluene given (1.55 g), and the molecular weight of toluene (92 g/mol).

(-3.91 x 10³ kJ/mol)(1 mol toluene / 92 g)(1.55g) = -65.875 kJ

So, the reaction is giving off 65.875 kJ of heat, and the calorimeter is absorbing that heat.

Now, we can use the heat capacity equation, with the positive q value of 65.875 kJ (because the calorimeter is absorbing this heat):

C = q/ΔT = (65.875 kJ)/(37.57°C - 23.12°C) = 4.56 kJ/°C