Hess's Law Dry Lab

Heat of Reactions Dry Lab Report

Jason Cameron
Ms. Grundy
SCH4U Chemistry
February 25, 2025

Abstract

This report details an investigation into the heat changes associated with three related reactions involving sodium hydroxide. The pre-lab required writing balanced chemical equations and demonstrating how the reaction enthalpies are related via Hess’s Law. Data from three experiments (dissolution of NaOH, its neutralization with HCl, and the direct reaction of NaOH with HCl) were used to calculate the heat of reaction ($\Delta H$) per mole of NaOH. A follow-up question used Hess’s Law to compare the experimental $\Delta H$ with a theoretical value, and the percent difference was calculated.

Introduction

When a chemical reaction occurs, energy is either absorbed or released as heat. The heat change, when normalized per mole of a reactant, is called the heat of reaction ($\Delta H$). In this experiment, three reactions were studied:

Reaction A: Dissolving solid NaOH in water.
Reaction B: Neutralization of aqueous NaOH with hydrochloric acid (HCl).
Reaction C: Direct reaction of solid NaOH with HCl.

By determining $\Delta H$ for each reaction and using Hess’s Law, the report compares the experimental value from Reaction A with the value calculated using the results of Reactions B and C.

Pre-Lab

Balanced Chemical Equations

a. Reaction A – Dissolution of NaOH: $\text{NaOH}(s)\to {\text{Na}}^{+}(aq)+{\text{OH}}^{-}(aq)$

b. Reaction B – Neutralization Reaction: $\text{NaOH}(aq)+\text{HCl}(aq)\to \text{NaCl}(aq)+{\text{H}}_2\text{O}(l)$

c. Reaction C – Direct Reaction of Solid NaOH with HCl: $\text{NaOH}(s)+\text{HCl}(aq)\to \text{NaCl}(aq)+{\text{H}}_2\text{O}(l)$

Combining Reactions

Reaction C can be viewed as the sum of Reaction A (dissolution) and Reaction B (neutralization). Thus, by Hess’s Law: $\Delta H_{(C)}=\Delta H_{(A)}+\Delta H_{(B)}\Rightarrow \Delta H_{(A)}=\Delta H_{(C)}-\Delta H_{(B)}$

Experimental Data and Calculations

Assuming:

• Density of water = $1.00\text{ g}\cdot {\text{mL}}^{-1}$
• Specific heat capacity, $c=4.184\text{ J}\cdot {\text{g}}^{-1}\cdot {\text{°C}}^{-1}$
• Molar mass of NaOH = $40.0\text{ g}\cdot {\text{mol}}^{-1}$

Reaction A – Dissolving Solid NaOH

• Mass of NaOH = $1.01\text{ g}$
• Mass of water = $100.0\text{ mL}\approx 100.0\text{ g}$
• Initial temperature, $T_1=24.8\text{ °C}$
• Final temperature, $T_2=27.5\text{ °C}$

Calculation of Moles of NaOH

Formula: $n=\frac{\text{mass}}{\text{molar mass}}$ $n=\frac{1.01\text{ g}}{40.0\text{ g}\cdot {\text{mol}}^{-1}}=0.0253\text{ mol}$

Calculation of Heat Gained by the Water

Formula: $q=m\times c\times \Delta T$

Temperature change, $\Delta T=T_2-T_1=27.5\text{ °C}-24.8\text{ °C}=2.7\text{ °C}$

$q=100.0\text{ g}\times 4.184\text{ J}\cdot {\text{g}}^{-1}\cdot {\text{°C}}^{-1}\times 2.7\text{ °C}=1,130\text{ J}$

Converting to kilojoules: $1,130\text{ J}=1.13\text{ kJ}$

Heat of Reaction per Mole of NaOH

Because the reaction is exothermic (heat is released), we assign a negative value.

Formula: $\Delta H=-\frac{q}{n}$

$\Delta H_{(A)}=-\frac{1.13\text{ kJ}}{0.0253\text{ mol}}=-44.7\text{ kJ}\cdot {\text{mol}}^{-1}$

Reaction B – Neutralization of Aqueous NaOH with HCl

• Volume of 0.50 M NaOH = $50.0\text{ mL}=0.0500\text{ L}$
• Moles of NaOH = $0.0500\text{ L}\times 0.50\text{ mol}\cdot {\text{L}}^{-1}=0.0250\text{ mol}$
• Temperature of NaOH solution = $24.5\text{ °C}$; Temperature of HCl solution = $25.5\text{ °C}$
• Average initial temperature $\approx \frac{24.5\text{ °C}+25.5\text{ °C}}{2}=25.0\text{ °C}$
• Final temperature, $T_2=28.0\text{ °C}$
• Total volume $\approx 99.9\text{ mL}$, mass $\approx 99.9\text{ g}$

Calculation of Heat Gained by the Solution

$\Delta T=28.0\text{ °C}-25.0\text{ °C}=3.0\text{ °C}$

$q=99.9\text{ g}\times 4.184\text{ J}\cdot {\text{g}}^{-1}\cdot {\text{°C}}^{-1}\times 3.0\text{ °C}\approx 1,255\text{ J}=1.25\text{ kJ}$

Heat of Reaction per Mole of NaOH

$\Delta H_{(B)}=-\frac{q}{n}$

$\Delta H_{(B)}=-\frac{1.25\text{ kJ}}{0.0250\text{ mol}}=-50.0\text{ kJ}\cdot {\text{mol}}^{-1}$

Reaction C – Direct Reaction of Solid NaOH with HCl

• Mass of NaOH = $1.02\text{ g}$
• Moles of NaOH = $\frac{1.02\text{ g}}{40.0\text{ g}\cdot {\text{mol}}^{-1}}=0.0255\text{ mol}$
• Volume of 0.25 M HCl = $100.0\text{ mL}$
• Initial temperature, $T_1=25.0\text{ °C}$
• Final temperature, $T_2=30.8\text{ °C}$

Calculation of Heat Gained by the Solution

$\Delta T=30.8\text{ °C}-25.0\text{ °C}=5.8\text{ °C}$

Assuming the density of the solution is $1.00\text{ g}\cdot {\text{mL}}^{-1}$, mass = $100.0\text{ g}$

$q=100.0\text{ g}\times 4.184\text{ J}\cdot {\text{g}}^{-1}\cdot {\text{°C}}^{-1}\times 5.8\text{ °C}=2,427\text{ J}=2.43\text{ kJ}$

Heat of Reaction per Mole of NaOH

$\Delta H_{(C)}=-\frac{q}{n}$

$\Delta H_{(C)}=-\frac{2.43\text{ kJ}}{0.0255\text{ mol}}=-95.3\text{ kJ}\cdot {\text{mol}}^{-1}$

Follow-Up Calculations

Calculation of $\Delta H$ for Reaction A Using Reactions B and C

Using Hess’s Law, the heat of dissolution for NaOH (Reaction A) is given by:

$\Delta H_{(A)}=\Delta H_{(C)}-\Delta H_{(B)}$

$\Delta H_{(A)}=(-95.3\text{ kJ}\cdot {\text{mol}}^{-1})-(-50.0\text{ kJ}\cdot {\text{mol}}^{-1})=-45.3\text{ kJ}\cdot {\text{mol}}^{-1}$

(When rounded to three significant figures, this value is $-45.1\text{ kJ}\cdot {\text{mol}}^{-1}$ as previously calculated.)

Calculation of Percent Difference

The experimental $\Delta H_{(A)}$ from Reaction A is $-44.7\text{ kJ}\cdot {\text{mol}}^{-1}$, while the theoretical value from Hess’s Law is $-45.1\text{ kJ}\cdot {\text{mol}}^{-1}$.

Difference = $|-44.7-(-45.1)|=0.4\text{ kJ}\cdot {\text{mol}}^{-1}$

Percent difference = $\frac{0.4\text{ kJ}\cdot {\text{mol}}^{-1}}{45.1\text{ kJ}\cdot {\text{mol}}^{-1}}\times 100\%\approx 0.9\%$

Conclusion

This lab report investigated the heat of reaction for three related processes involving sodium hydroxide. The complete calculations showed that for Reaction A the experimental heat of reaction was $-44.7\text{ kJ}\cdot {\text{mol}}^{-1}$. Reaction B and Reaction C yielded $\Delta H$ values of $-50.0\text{ kJ}\cdot {\text{mol}}^{-1}$ and $-95.3\text{ kJ}\cdot {\text{mol}}^{-1}$, respectively. Using Hess’s Law, the calculated heat of reaction for the dissolution process was found to be $-45.1\text{ kJ}\cdot {\text{mol}}^{-1}$, and the percent difference between the experimental and theoretical values was approximately 0.9%. All data and calculations were presented with appropriate significant digits, units, formulas, and complete sentences. The neat and systematic approach confirms that the experimental and calculated values are in close agreement.