A 2.2 g sample of quinone #C_6H_4O_2# was burned in a bomb calorimeter for which the total heat capacity is 7854 J/°C. The temperature of the calorimeter increased from 23.44 C to 30.57 °C. What is the molar heat of combustion of quinone (in kJ/mol)?

1 Answer
Jan 3, 2017

Answer:

The molar heat of combustion of quinone is -2700 kJ/mol.

Explanation:

There are two heat transfers involved.

#"heat of combustion of quinone + heat gained by calorimeter = 0"#

#q_1 + q_2 = 0#

#nΔ_ cH + C_"cal"ΔT = 0#

In this problem,

#n = 2.2 color(red)(cancel(color(black)("g quinone"))) × "1 mol quinone"/(108.09 color(red)(cancel(color(black)("g quinone")))) = "0.0204 mol quinone"#

#C_"cal" = "7854 J°C"^"-1"#

#ΔT = T_f - T_i = "30.57 °C - 23.44 °C" = "7.13 °C"#

#q_1 = nΔ_cH = 0.0204 Δ_cH color(white)(l)"mol"#

#q_2 = C_"cal"ΔT = "7854 J" color(red)(cancel(color(black)("°C"^"-1"))) × 7.13 color(red)(cancel(color(black)("°C"))) = "56 000 J" = "56.0 kJ"#

#q_1 + q_2 =0.0204 Δ_cHcolor(white)(l) "mol" + "56.0 kJ" = 0#

#Δ_cH = ("-56.0 kJ")/("0.0204 mol") = "-2700 kJ/mol"# (2 significant figures)