Octane. C_sH_1 s. a primary constituent of gasoline, bums in air C_8H_18(9+ 12 1
ID: 1008469 • Letter: O
Question
Octane. C_sH_1 s. a primary constituent of gasoline, bums in air C_8H_18(9+ 12 1/2o_2(g)- 8 CO_2(g)- 9 H_20(C) A 1.00-g sample of octane is burned in a constant volume calorimeter. The calorimeter is in an insulated container with 1.20 kg of water. the temperature of the water and the bomb rises from 25.00 degree C (298.15 K) to 33.21 degree C (306.36 K). The heat required to raise the bomb's temperature (its heat capacity). C_bomb- is 837 J/K. The specific heat of water is 4.1S4 J/g-K. What is the heat of combustion per gram of octane? What is the heat of combustion per mole of octane?Explanation / Answer
Heat released in reaction = Heat absorbed by solution + heat absorbed by calorimeter
water mass = 1.2 kg = 1200 g
octane mass = 1g , solution mass = 1200+1 = 1201 g
Temperature change = final Temp - initial temp = 33.21 - 25 = 8.21
we have formula
Heat absorbed by solution = specific heat of solution x temp change x mass of solution
= 4.184 J/gK x 8.21 x 1201
= 47467 J = 47.464 KJ
Heat absorbed by calorimeter = specific heat of calorimeter x temp change
= 837 J/K x 8.21 K
= 6871.77 J = 6.87 KJ
total heat absobed = 47.464 + 6.87 = 54.336 KJ
Moles of octane = mass/molar mass of octane = 1/ 114.23 = 0.008754
now Heat of combustion per g octane = - 54.336 KJ/g ( -ve sign indicates heat released in reaction)
Heat of combustion per mol = -54.336 /0.008754 = -6207 KJ /mol
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.