Question #2e5cf

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Question #2e5cf

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Answer 1 · 2017-06-27T00:08:41

Warning! Long Answer.
The heat capacity of the calorimeter is 1.361 kJ/°C.
The enthalpy of combustion of propan-2-ol is -1985 kJ/mol.

Explanation:

This is a two-part problem.

Step 1. Calculate the heat capacity of the calorimeter

There are two heat flows involved in this problem.

$Heat from combustion of methanol + heat absorbed by calorimeter = 0$ $"Heat from combustion of methanol + heat absorbed by calorimeter" = 0$

$m m m m m m m q_{1} m m m m m m m m m + m m m m m m m q_{2} m m m m m m = 0$ $color(white)(mmmmmmm)q_1 color(white)(mmmmmmmmm)+ color(white)(mmmmmmm) q_2color(white)(mmmmmm) = 0$

$color(white)(mmmmmm)nΔ_text(c)H color(white)(mmmmmmmm) + color(white)(mmmmm)C_text(Cal)ΔT color(white)(mmmmm) = 0$

Let's calculate these two terms separately.

$n = 2.00 color(red)(cancel(color(black)("g MeOH"))) × " 1 mol MeOH"/(32.04 color(red)(cancel(color(black)("g MeOH")))) = "0.062 42 mol MeOH"$

$q_1 = nΔ_text(c)H = "0.062 42" color(red)(cancel(color(black)("mol MeOH"))) × "-715 kJ"/(1 color(red)(cancel(color(black)("mol MeOH")))) = "-44.63 kJ"$

$ΔT = "(52.4 - 19.6) °C" = "32.8 °C"$

$q_2 = C_text(Cal)ΔT = C_text(Cal) × "32.8 °C" = 32.8C_text(Cal) color(white)(l)"°C"$

$q_1 + q_2 = 0$

$"-44.63 kJ" + 32.8C_text(Cal) color(white)(l)"°C" = 0$

$C_text(Cal) = "44.63 kJ"/"32.8 °C" = "1.361 kJ/°C"$

Step 2. Use the heat capacity of the calorimeter to calculate the heat of combustion of propan-2-ol

As before, there are two heat flows involved.

$"Heat from combustion of propan-2-ol + heat absorbed by calorimeter" = 0$

$color(white)(mmmmmmmm)q_1 color(white)(mmmmmmmmm)+ color(white)(mmmmmmm) q_2color(white)(mmmmmm) = 0$

$color(white)(mmmmmmm)nΔ_text(c)H color(white)(mmmmmmmm) + color(white)(mmmmm)C_text(Cal)ΔT color(white)(mmmmm) = 0$

Let's calculate these two terms separately.

$n = 1.50 color(red)(cancel(color(black)("g PrOH"))) × " 1 mol PrOH"/(60.10 color(red)(cancel(color(black)("g PrOH")))) = "0.024 96 mol PrOH"$

$q_1 = nΔ_text(c)H = "0.024 96 mol" × nΔ_text(c)H = "0.024 96" Δ_text(c)H color(white)(l)"mol"$

$ΔT = "(56.2 - 19.8) °C" = "36.4 °C"$

$q_2 = C_text(Cal)ΔT = "1.361 kJ"·color(red)(cancel(color(black)("°C"^"-1"))) × 36.4 color(red)(cancel(color(black)("°C"))) = "49.53 kJ"$

$q_1 + q_2 = 0$

$"0.024 96" Δ_text(c)H color(white)(l)"mol" + "49.53 kJ" = 0$

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