Question

Question #9adde

Answer 1

You have to subtract the heat capacity term for the calorimeter because the heat of reaction warms both the water and the calorimeter.

There are three heats to consider.

heat of reaction + heat to warm water + heat to warm calorimeter = 0

`q_1 + q_2 + q_3` = 0

`nΔH_"rxn" + mcΔT + C_"Cal"ΔT` = 0

So

`nΔH_"rxn" = -mcΔT - C_"Cal"ΔT`

This shows that you have to subtract the heat capacity term for the calorimeter when you are calculating the heat of reaction.

EXAMPLE

A student reacts 0.494 g of Mg with 100 g of dilute HCl in a calorimeter with a heat capacity of 133 J/°C. The temperature of the solution changes from 25.0 °C to 41.8°C. The equation for the reaction is Mg + 2HCl → MgCl₂ + H₂. Assume that the specific heat capacity of the solution is 4.18 J·g⁻¹°C⁻¹. What is the heat of reaction?

Solution

`nΔH_"rxn" = -mcΔT - C_"Cal"ΔT`

`n` = 0.494 g Mg × `(1"mol Mg")/(24.30"g Mg")` = 0.020 33 mol Mg (3 significant figures + 1 guard digit)

`ΔT` = (41.8 – 25.0) °C = 16.8 °C

So

0.020 33 mol × `ΔH_"rxn"` = -(100 g × 4.18 J·g⁻¹°C⁻¹ × 16.8 °C) – (133 J·°C⁻¹ × 16.8 °C) = -7022 J – 2223 J = -9256 J

`ΔH_"rxn" = (-9256"J")/(0.020 33"mol")` = -455 000 J/mol = -455 kJ/mol (3 significant figures)