Question

Question #55fec

Answer 1

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"`