Question #2e5cf

1 Answer
Jun 27, 2017

Answer:

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#

#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#

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