Question

A 21.8 g sample of ethanol (C₂H₅OH, 46.07 g/mol) is burned in a bomb calorimeter, according to the following reaction equation. If the temperature of the rises from 13.0 °C to 79.3 °C, what is the heat capacity of the calorimeter?

C₂H₅OH(l) + 3 O₂(g) →2 CO₂(g) + 3 H₂O(g)    ΔH°_rxn = –1235 kJ

  

Answer 1

Given, the balanced chemical reaction,

C₂H₅OH(l) + 3O₂(g) 2CO₂(g) + 3H₂O(g) --------- H^o_rxn = -1235 kJ

Also given,

Mass of ethanol(C₂H₅OH) = 21.8 g

Molar mass of ethanol = 46.07 g/mol

Initial temperature(T_i) = 13.0 ^oC

Final temperature(T_f) = 79.3 ^oC

Calculating the number of moles of ethanol from the given mass and molar mass,

= 21.8 g C₂H₅OH x ( 1 mol / 46.07 g)

= 0.4732 mol ethanol

From the reaction, Heat released = 1235 kJ / mol of C₂H₅OH.

Now, the heat release for 0.4732 mol ethanol is,

= 0.4732 mol ethanol x ( 1235 kJ / 1 mol ethanol)

= 584.4 kJ

Now, Temperature difference(T) = T_f - T_i

T = 79.3 ^oC - 13.0 ^oC

T = 66.3 ^oC

Thus, Heat capacity of bomb calorimeter is,

= Heat released by 21.8 g sample of ethanol / T

= 584.4 kJ / 66.3 ^oC

= 8.81 kJ/ ^oC