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# Chapter 6: Complete Guide To Thermochemistry

Thermochemistry is the study covering the heat absorbed or evolved during chemical reactions.

Energy and Its Units

Energy can be defined as the potential or capacity to move matter (or do work)

Kinetic Energy: energy due to the motion of an object

Potential Energy: energy due to position of an object

Chemical Energy: energy stored inside chemical substances

law of conservation of energy states that energy may be converted from one form to another, but the total quantity of energy remains constant.

The unit of energy is Joule (J)

1J = 1 kg*m^2/s^2

Let's define a couple more terms.

System = substance or mixture of substances we are studying

Surroundings = Everything else in the universe

Heat is defined as the energy that flows into or out of a system due to the difference in temperature between the system and its surroundings.

ΔE = q+w where E is the internal energy of the system, q is heat and w is work.

We can think of system as being lazy and greedy. When it has to do work, work is negative and when the system releases heat, heat is negative.

Enthalpy and Enthalpy Change

Enthalpy (H) is a property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction.

Enthalpy is a state function (a property of a system that depends only on its present state and is independent of any previous history of the system.)

Under constant pressure,ΔH = qp (The enthalpy of reaction equals the heat of reaction at constant pressure.)

Applying Stoichiometry to Heats of Reaction

We can use the coefficients in a balanced equation to calculate the heat released or absorbed when we are given a specific amount of the reactant or product.

ΔH for a reaction represents the amount of heat per mol or reaction.

For example: 2A+3B --> C ΔH=20kJ/mol

This means 20kJ of heat are absorbed when 2 moles of A react with 3 moles of B to produce 1 mol of C.

Problem: How much heat is evolved when 9.07 * 10^5 g of ammonia is produced according to the following equation? (Assume that the reaction occurs at constant pressure.)

N2(g) + 3H2(g)--> 2NH3(g); ΔH = -91.8 kJ

The ΔH that is given is per 1 mole of the reaction. We can use coefficients in the equation to relate it to ΔH. So, 91.8 kJ of heat are released for 1 mole of N2, 3 moles of H2 and 2 moles of NH3.

First, we need to convert grams to moles using molar mass found in the Periodic Table.

9.07 * 10^5 g *(1mol/17.0 g ) = 53353.941mol NH3

Now, we can make a ratio and solve for the heat.

91.8 kJ = x

2 mol NH3( from the equation) 53353.941mol NH3

x=2.45 * 10^6 kJ of heat evolves

Measuring Heats of Reaction (Calorimetry)

C = heat capacity = is the amount of heat needed to raise the temperature of the sample of substance one degree Celsius (or one kelvin)

q=CΔT

c = specific heat capacity = amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one kelvin) at constant pressure.

q=mcΔT where q=heat (J), m=mass (g) and c is specific heat capacity (J/gC)

Example: Calculate the heat absorbed by 15.0 g of water to raise its temperature from 20.0°C to 50.°0C (at constant pressure). The specific heat of water is 4.18 J/(g*°C).

q=mcΔT

q= (15.0 g)(4.18 J/(g*C))(50.0°C-20.0°C) = 1.88 * 10^3 J

Calorimetry is the measurement of heat produced or absorbed during a process.

Calorimetry problems will generally consist of two or three parts:

Reaction : for reaction we will use q=mcΔT

Water: q=mcΔT

Calorimeter (container where the reaction takes place): q=CΔT

The heats of all the parts will add up to 0.

qrxn+qwater+qcal = 0

Note: There may be other calorimetry problems not involving water or reaction.

Constant-Volume Calorimetry

qsystem =0

Constant- Pressure Calorimetry

qrxn = ΔH

Calorimetry Problem: Suppose 0.562 g of graphite is placed in a calorimeter with an excess of oxygen at 25.00°C and 1 atm pressure . Excess O2 ensures that all carbon burns to form CO2.

During the reaction, the calorimeter temperature rises from 25.00°C to 25.89°C. The heat capacity of the calorimeter and its contents was determined in a separate experiment to be 20.7 kJ/C. What is the heat of reaction at 25.00°C and 1 atm pressure (ΔH) ?

C(graphite) + O2(g) --> CO2(g)

We first determine all the parts of the experiment. We have the reaction (burning of graphite) as well as the calorimeter (container).

qrxn+qcal = 0

qrxn = -qcal

qcal = CΔT

qrxn = -CcalΔT = -20.7 kJ/°C * (25.89°C - 25.00°C) = -20.7 kJ/°C * 0.89°C = -18.4 kJ

The heat we got is for burning 0.562 g of graphite. We need to convert grams to moles and then divide 18.4 kJ by the moles to determine heat released per mol of graphite.

(0.562 g of C/ 12gC )=0.04683 mol

-18.4 kJ/0.04683 molC = -3.9 * 10^2 kJ/molC

Hess’s Law

Hess's Law is another way to determine ΔH of the reaction. It states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps.

You will recognize Hess's law when you will be given multiple equations with their ΔH's and will be asked for a ΔH for a specific reaction.

Let's first go over some rules:

1. Reverse equation = Change the sign of ΔH

2. Multiply equation by a number = Multiply ΔH by the same number

Example: What is the enthalpy of reaction, ΔH, for the formation of tungsten carbide, WC, from the elements? Final: W(s) + C(graphite) --> WC(s)

1. 2W(s) + 3O2(g) --> 2WO3(s); ΔH = -1685.8 kJ

2. C(graphite) + O2(g) --> CO2(g); ΔH = -393.5 kJ

3. 2WC(s) + 5O2(g) --> 2WO3(s) + 2CO2(g); ΔH = -2391.8 kJ

Mayya's Trick: Find a substance in an equation that it has in common with the final equation we are trying to find ΔH for but not in common with the rest of steps given.

For example, equation 1 has W(s) which equations 2 and 3 do not have but the final equation does. In step 1 we see 2 in front of W(s) but in the final equation we do not see a number (signifying a coefficient of 1). This means we need to multiply equation 1 and its ΔH by 1/2

(2W(s) + 3O2(g) --> 2WO3(s))(1/2); ΔH = -1685.8 kJ(1/2) = -842.9 kJ

Equation 2 has C(graphite) in common with the final equation but not other steps. It has the same coefficient and is on the same side as in the final equation. We leave equation 2 as is.

C(graphite) + O2(g) --> CO2(g); ΔH = -393.5 kJ

Equation 3 has WC(s) in common with the final equation but not with the other steps. It goes from coefficient of 2 to coefficient of 1 (final equation) which means we need to multiply it by 1/2. Also, it is on the reactant (left) side while in the final equation it is on the product side (right). We need to reverse the equation as well. When we reverse an equation, we need to change the sign of ΔH.

After reversing and multiply by 1/2, we get:

WO3(s) + CO2(g) --> WC(s) + 5/2O2(g) ΔH = +2391.8 kJ/2 = 1195.9 kJ

Now, lets add all the steps to see if they add up to the final equation.

1. W(s) + 3/2O2(g) --> WO3(s)

2. C(graphite) + O2(g) --> CO2(g)

3. WO3(s) + CO2(g) --> WC(s) + 5/2O2(g)

We can cross out the substances that we see on both sides such as CO2(g) and we get W(s) + C(graphite) --> WC(s) which is the same as the final equation.

To get the ΔH for the final equation we just add up ΔHs for the steps. We get:

ΔH = ΔH1+ΔH2+ΔH3 = -40.5 kJ

Standard Enthalpies of Formation

Standard enthalpy of formation of a substance (ΔH°f, is the enthalpy change for the formation of one mole of the substance in its standard state (1 atm pressure and the specified temperature (usually 25°C). from its elements in their reference form and in their standard states.

Using ΔH°f of the reactants and products we can calculate ΔH of the reaction.

Example: 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g)

What is the standard enthalpy change for this reaction?

Solution: We need to look up ΔH°f for the products and reactants and plug them into our formula.

ΔH° = [4(90.3) + 6(-241.8)] kJ - [4(-45.9) + 5(0)] kJ

= -906 kJ

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