Question

It is known that the combustion of 1.00 g of benzoic acid produces 26.4 kJ of heat.

(i) Balance the following combustion equation using the lowest possible whole numbers as coefficients.
C6H5COOH(s) + O2(g) → CO2(g) + H2O(l)
(ii) Calculate the change in enthalpy, ΔH, for the balanced equation written in part (c)(i). Show your work below, and express your answer in units of kJ/molrxn. Make sure to include the sign of ΔH in your answer.
(iii) In a certain experiment, 2.25 g of benzoic acid was burned completely in a bomb calorimeter. Calculate the amount of heat released from the combustion of 2.25 g of benzoic acid.
(iv) Assuming that all of the heat generated from the combustion of 2.25 g of benzoic acid is completely transferred to a 500. g sample of water (specific heat = 4.18 J/(g·oC)) at 25.0oC, calculate the final temperature you would expect to observe in this sample of water after it is heated

Answer 1

i)

The balanced equation is $\ C_6H_5COOH(s) +\dfrac{15}{2} O_2(g) → 7CO_2(g) +3 H_2O(l)$

ii)

$Molecular \ mass\ of\ benzoic\ acid=12*6+5+12+16*2+1=122 g$

as 1 g produces 24.6 kJ of heat so heat produced by one mole of benzoic acid is equal to

=24.6*122 =3001.2 kJ which is the $\Delta H$

iii) Heat released from 2.25 g of benzoic acid = 2.25*24.6 = 55.35 kJ

iv) change in heat = ms (T₂-T₁)

55.35*10³ =500*4.18*(T₂-20)

T₂ =46.5^oC