Calculate The Heat Of Reaction Δh For The Following Reaction

Diving into the world of thermochemistry, understanding how to calculate the heat of reaction (ΔH) is crucial for chemists, engineers, and anyone interested in predicting the energy changes that accompany chemical reactions. This article will provide a comprehensive guide on calculating ΔH, walking you through various methods, and illustrating them with practical examples.

Understanding the Heat of Reaction (ΔH)

The heat of reaction, also known as enthalpy change (ΔH), represents the amount of heat absorbed or released during a chemical reaction at constant pressure. It's a fundamental concept in thermodynamics that helps us understand whether a reaction is exothermic (releases heat, ΔH < 0) or endothermic (absorbs heat, ΔH > 0). The value of ΔH is typically expressed in kilojoules per mole (kJ/mol).

Before we delve into the methods of calculation, let's clarify some key terms:

• System: The specific part of the universe that is under investigation. In chemistry, this is typically the reaction itself.
• Surroundings: Everything else outside the system.
• Enthalpy (H): A thermodynamic property of a system that is the sum of its internal energy and the product of its pressure and volume. Enthalpy is a state function, meaning its value depends only on the initial and final states, not the path taken.
• Standard Enthalpy Change (ΔH°): The enthalpy change when a reaction is carried out under standard conditions (298 K and 1 atm pressure).

Methods for Calculating the Heat of Reaction (ΔH)

There are several methods available to calculate the heat of reaction, each with its own advantages and limitations. We will discuss the following methods:

1. Using Standard Enthalpies of Formation
2. Using Hess's Law
3. Using Calorimetry
4. Using Bond Energies

1. Using Standard Enthalpies of Formation (ΔH°f)

The standard enthalpy of formation (ΔH°f) is the change in enthalpy when one mole of a substance is formed from its elements in their standard states under standard conditions (298 K and 1 atm). The standard state for an element is its most stable form under standard conditions (e.g., O2(g) for oxygen, C(s, graphite) for carbon).

The heat of reaction can be calculated using the following formula:

ΔH°reaction = Σ ΔH°f(products) - Σ ΔH°f(reactants)

Where:

• Σ ΔH°f(products) is the sum of the standard enthalpies of formation of all products, each multiplied by its stoichiometric coefficient in the balanced chemical equation.
• Σ ΔH°f(reactants) is the sum of the standard enthalpies of formation of all reactants, each multiplied by its stoichiometric coefficient in the balanced chemical equation.

Important Notes:

• The standard enthalpy of formation of an element in its standard state is defined as zero (e.g., ΔH°f[O2(g)] = 0, ΔH°f[C(s, graphite)] = 0).
• Standard enthalpy of formation values are typically found in thermodynamic tables or databases.

Example:

Calculate the standard enthalpy change for the combustion of methane (CH4):

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Given the following standard enthalpies of formation:

• ΔH°f[CH4(g)] = -74.8 kJ/mol
• ΔH°f[O2(g)] = 0 kJ/mol
• ΔH°f[CO2(g)] = -393.5 kJ/mol
• ΔH°f[H2O(g)] = -241.8 kJ/mol

Using the formula:

ΔH°reaction = [ΔH°f(CO2(g)) + 2 * ΔH°f(H2O(g))] - [ΔH°f(CH4(g)) + 2 * ΔH°f(O2(g))]

ΔH°reaction = [(-393.5 kJ/mol) + 2 * (-241.8 kJ/mol)] - [(-74.8 kJ/mol) + 2 * (0 kJ/mol)]

ΔH°reaction = (-393.5 - 483.6) - (-74.8)

ΔH°reaction = -877.1 + 74.8

ΔH°reaction = -802.3 kJ/mol

Therefore, the combustion of methane is an exothermic reaction, releasing 802.3 kJ of heat per mole of methane combusted.

2. Using Hess's Law

Hess's Law states that the enthalpy change for a reaction is independent of the pathway taken. In other words, if a reaction can be carried out in a series of steps, the sum of the enthalpy changes for each step will be equal to the enthalpy change for the overall reaction.

This law is particularly useful when it is difficult or impossible to measure the enthalpy change directly. By manipulating known enthalpy changes of other reactions, we can calculate the enthalpy change for the target reaction.

Key Steps in Applying Hess's Law:

1. Identify the target reaction: This is the reaction for which you want to calculate the enthalpy change.
2. Find a series of reactions (intermediate reactions) that, when added together, give the target reaction. These intermediate reactions must have known enthalpy changes.
3. Manipulate the intermediate reactions:
   • If an intermediate reaction needs to be reversed to get the reactants and products on the correct sides, change the sign of its enthalpy change (ΔH becomes -ΔH).
   • If an intermediate reaction needs to be multiplied by a coefficient to match the stoichiometric coefficients in the target reaction, multiply its enthalpy change by the same coefficient.
4. Add the manipulated intermediate reactions together: Make sure that all species that appear on both sides of the equation cancel out, leaving you with the target reaction.
5. Add the manipulated enthalpy changes together: The sum of these enthalpy changes is the enthalpy change for the target reaction.

Example:

Calculate the enthalpy change for the following reaction:

2C(s, graphite) + O2(g) → 2CO(g)

Given the following reactions and their enthalpy changes:

1. C(s, graphite) + O2(g) → CO2(g) ΔH1 = -393.5 kJ/mol
2. 2CO(g) + O2(g) → 2CO2(g) ΔH2 = -566.0 kJ/mol

Solution:

1. Target Reaction: 2C(s, graphite) + O2(g) → 2CO(g)

2. Intermediate Reactions: Reactions 1 and 2 above.

3. Manipulation:

   • Multiply reaction 1 by 2: 2C(s, graphite) + 2O2(g) → 2CO2(g) ΔH1' = 2 * (-393.5 kJ/mol) = -787.0 kJ/mol
   • Reverse reaction 2 and divide by 2 (to get 1 mole of O2 on the reactant side): CO2(g) → CO(g) + 1/2 O2(g) ΔH2' = (1/2) * (+566.0 kJ/mol) = +283.0 kJ/mol
   • Multiply the reversed reaction by 2 to match stoichiometric coefficients in the target reaction: 2CO2(g) → 2CO(g) + O2(g) ΔH2'' = +566.0 kJ/mol

4. Add the manipulated reactions:

   2C(s, graphite) + 2O2(g) → 2CO2(g) ΔH1' = -787.0 kJ/mol

   2CO2(g) → 2CO(g) + O2(g) ΔH2'' = +566.0 kJ/mol

   Adding the reactions:

   2C(s, graphite) + 2O2(g) + 2CO2(g) → 2CO2(g) + 2CO(g) + O2(g)

   Simplifying:

   2C(s, graphite) + O2(g) → 2CO(g) (Target Reaction)

5. Add the manipulated enthalpy changes:

   ΔH°reaction = ΔH1' + ΔH2'' = -787.0 kJ/mol + 566.0 kJ/mol = -221.0 kJ/mol

Therefore, the enthalpy change for the reaction 2C(s, graphite) + O2(g) → 2CO(g) is -221.0 kJ/mol.

3. Using Calorimetry

Calorimetry is the experimental measurement of heat transfer during a chemical or physical process. A calorimeter is a device used to measure the heat absorbed or released.

The basic principle of calorimetry is based on the conservation of energy: the heat released or absorbed by the system (reaction) is equal to the heat absorbed or released by the surroundings (calorimeter and its contents).

Types of Calorimeters:

• Constant-Volume Calorimeter (Bomb Calorimeter): Used for reactions involving gases or reactions where volume changes are negligible. The reaction is carried out in a sealed vessel (bomb) at constant volume. The heat released or absorbed is measured by monitoring the temperature change of the surrounding water.
• Constant-Pressure Calorimeter (Coffee-Cup Calorimeter): A simple calorimeter made from two nested coffee cups. This type is used for reactions carried out in solution at constant pressure (atmospheric pressure). The heat released or absorbed is measured by monitoring the temperature change of the solution.

Calculations:

The heat transferred (q) is calculated using the following equation:

q = mcΔT

Where:

• q is the heat transferred (in joules or kilojoules)
• m is the mass of the substance that is changing temperature (in grams or kilograms)
• c is the specific heat capacity of the substance (in J/g°C or kJ/kg°C)
• ΔT is the change in temperature (in °C or K)

For a reaction in a calorimeter:

qreaction = -qcalorimeter

For a constant-volume calorimeter (bomb calorimeter):

qcalorimeter = Ccalorimeter * ΔT

Where Ccalorimeter is the heat capacity of the calorimeter.

For a constant-pressure calorimeter (coffee-cup calorimeter):

qcalorimeter = mcΔT

Where m and c refer to the mass and specific heat capacity of the solution in the calorimeter.

Example:

50.0 mL of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH in a coffee-cup calorimeter. The initial temperature of both solutions is 22.0 °C. After mixing, the temperature rises to 28.9 °C. Assume the density of the solution is 1.0 g/mL and the specific heat capacity of the solution is 4.184 J/g°C. Calculate the heat of reaction.

Solution:

1. Calculate the total mass of the solution:

   Total volume = 50.0 mL + 50.0 mL = 100.0 mL

   Mass = Volume * Density = 100.0 mL * 1.0 g/mL = 100.0 g

2. Calculate the temperature change:

   ΔT = Tfinal - Tinitial = 28.9 °C - 22.0 °C = 6.9 °C

3. Calculate the heat absorbed by the solution (qcalorimeter):

   qcalorimeter = mcΔT = (100.0 g) * (4.184 J/g°C) * (6.9 °C) = 2886.96 J = 2.887 kJ

4. Calculate the heat of reaction (qreaction):

   qreaction = -qcalorimeter = -2.887 kJ

5. Calculate the moles of HCl (or NaOH) reacted:

   Moles = Volume * Molarity = (0.050 L) * (1.0 mol/L) = 0.050 mol

6. Calculate the enthalpy change per mole (ΔH):

   ΔH = qreaction / moles = -2.887 kJ / 0.050 mol = -57.74 kJ/mol

Therefore, the heat of reaction for the neutralization of HCl with NaOH is -57.74 kJ/mol.

4. Using Bond Energies

Bond energy is the average energy required to break one mole of a particular bond in the gaseous phase. Bond energies are always positive values because energy is always required to break a bond.

The heat of reaction can be estimated using bond energies with the following formula:

ΔH ≈ Σ Bond energies(reactants) - Σ Bond energies(products)

Where:

• Σ Bond energies(reactants) is the sum of the bond energies of all bonds broken in the reactants.
• Σ Bond energies(products) is the sum of the bond energies of all bonds formed in the products.

Important Notes:

• This method provides an estimation of the heat of reaction because it uses average bond energies. The actual bond energy can vary depending on the specific molecule.
• This method is most accurate for reactions in the gaseous phase.
• Make sure to draw the Lewis structures of the reactants and products to correctly identify all the bonds present.

Example:

Estimate the enthalpy change for the following reaction:

H2(g) + Cl2(g) → 2HCl(g)

Given the following average bond energies:

• H-H bond: 436 kJ/mol
• Cl-Cl bond: 242 kJ/mol
• H-Cl bond: 431 kJ/mol

Solution:

1. Identify the bonds broken and formed:

   • Reactants: 1 H-H bond and 1 Cl-Cl bond are broken.
   • Products: 2 H-Cl bonds are formed.

2. Calculate the sum of bond energies for reactants:

   Σ Bond energies(reactants) = (1 * 436 kJ/mol) + (1 * 242 kJ/mol) = 678 kJ/mol

3. Calculate the sum of bond energies for products:

   Σ Bond energies(products) = (2 * 431 kJ/mol) = 862 kJ/mol

4. Calculate the estimated enthalpy change:

   ΔH ≈ Σ Bond energies(reactants) - Σ Bond energies(products) = 678 kJ/mol - 862 kJ/mol = -184 kJ/mol

Therefore, the estimated enthalpy change for the reaction H2(g) + Cl2(g) → 2HCl(g) is -184 kJ/mol.

Factors Affecting the Heat of Reaction

Several factors can influence the heat of reaction, including:

• Temperature: Enthalpy changes are temperature-dependent, although the dependence is usually small unless there is a phase change or a significant change in heat capacity.
• Pressure: Enthalpy changes are also pressure-dependent, but the effect is usually small for reactions involving only condensed phases (liquids and solids). For reactions involving gases, the effect can be more significant.
• Physical States of Reactants and Products: The physical states (solid, liquid, gas) of the reactants and products can significantly affect the enthalpy change. Phase changes involve energy input or release, so the enthalpy change will be different if a substance is a gas versus a liquid.
• Concentration: For reactions in solution, the concentration of the reactants can affect the heat of reaction, particularly for reactions that are not elementary (i.e., reactions that occur in multiple steps).
• Presence of Catalysts: Catalysts do not affect the overall enthalpy change of a reaction. They only affect the rate of the reaction by lowering the activation energy.

Applications of Heat of Reaction

Understanding and calculating the heat of reaction has numerous applications in various fields:

• Chemical Engineering: Designing and optimizing chemical reactors requires accurate knowledge of the enthalpy changes of the reactions involved. This information is crucial for controlling temperature, preventing explosions, and maximizing product yield.
• Materials Science: Predicting the stability and reactivity of materials often involves calculating the enthalpy changes of relevant reactions. This is important for developing new materials with desired properties.
• Environmental Science: Understanding the enthalpy changes of combustion reactions is crucial for assessing the environmental impact of burning fuels and designing cleaner energy technologies.
• Biochemistry: Biochemical reactions involve energy changes that are vital for life processes. Understanding the enthalpy changes of these reactions is important for studying metabolism, enzyme kinetics, and other biological phenomena.
• Everyday Life: Understanding exothermic and endothermic reactions can help explain phenomena such as why ice melts (endothermic) or why burning wood releases heat (exothermic).

Conclusion

Calculating the heat of reaction (ΔH) is a fundamental skill in chemistry with wide-ranging applications. By mastering the different methods – using standard enthalpies of formation, Hess's Law, calorimetry, and bond energies – you can gain a deeper understanding of the energy changes that accompany chemical reactions. Each method has its own strengths and weaknesses, and the choice of method depends on the available data and the desired level of accuracy. A solid grasp of these concepts will empower you to predict, analyze, and manipulate chemical reactions more effectively.