Question #55fec

1 Answer
Ernest Z.
Aug 17, 2017

The molar enthalpy of combustion of sulfur is -146 kJ/mol.

There are two heat transfers involved.

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

#q_1 + q_2 = 0#

#nΔ_ cH + mCΔT = 0#

In this problem,

#n = 64 color(red)(cancel(color(black)("g S"))) × "1 mol S"/(32.06 color(red)(cancel(color(black)("g S")))) = "2.00 mol S"#

#m = "1000 g"#

#C = "4.184 J·g"^"-1""°C"^"-1"#

#ΔT = T_f - T_i = "90 °C - 20 °C" = "70 °C"#

#q_1 = nΔ_cH = 2.00 Δ_cHcolor(white)(l) "mol"#

#q_2 = mCΔT = 1000 color(red)(cancel(color(black)("g"))) × "4.184 J"color(red)(cancel(color(black)("g"^"-1""°C"^"-1"))) × 70 color(red)(cancel(color(black)("°C"))) = "293 000 J" = "293 kJ"#

#q_1 + q_2 =2.00 Δ_cHcolor(white)(l) "mol" + "293 kJ" = 0#

#Δ_cH = ("-293 kJ")/("2.00 mol") = "-146 kJ/mol"#

Related questions
See all questions in Enthalpy
Impact of this question
1380 views around the world
You can reuse this answer
Creative Commons License