Question #55fec

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Question #55fec

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Answer 1 · 2017-08-17T01:35:18

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

There are two heat transfers involved.

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

$q_{1} + q_{2} = 0$ $q_1 + q_2 = 0$

$nΔ_ cH + mCΔT = 0$

In this problem,

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

$m = "1000 g"$

$C = "4.184 J·g"^"-1""°C"^"-1"$

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

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

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

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

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

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Question #55fec

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