# Question 2e5cf

Jun 27, 2017

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.

$\text{Heat from combustion of methanol + heat absorbed by calorimeter} = 0$

$\textcolor{w h i t e}{m m m m m m m} {q}_{1} \textcolor{w h i t e}{m m m m m m m m m} + \textcolor{w h i t e}{m m m m m m m} {q}_{2} \textcolor{w h i t e}{m m m m m m} = 0$

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"

${q}_{1} + {q}_{2} = 0$

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

${C}_{\textrm{C a l}} = \text{44.63 kJ"/"32.8 °C" = "1.361 kJ/°C}$

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.

$\text{Heat from combustion of propan-2-ol + heat absorbed by calorimeter} = 0$

$\textcolor{w h i t e}{m m m m m m m m} {q}_{1} \textcolor{w h i t e}{m m m m m m m m m} + \textcolor{w h i t e}{m m m m m m m} {q}_{2} \textcolor{w h i t e}{m m m m m m} = 0$

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"

${q}_{1} + {q}_{2} = 0$

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

Δ_text(c)H = "-49.54 kJ"/"0.024 96 mol" = "-1985 kJ·mol"^"-1""#