Heat Changes in Chemical Processes

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Kelley Dickerson
5 years ago
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1 Heat Changes in Chemical Processes 1. Zeroeth Law of Thermodynamics: Two systems at thermal equilibrium with a third system are in thermal equilibrium with each other. Paraphrase: Systems in thermal equilibrium with each other do not undergo a temperature change; there is no net exchange of heat at thermal equilibrium.

2 2. First Law of Thermodynamics: The change in the internal energy of a system equals the heat added to the system minus the work done by the system: U = Q - W Paraphrase: Energy is conserved. It can t be created or destroyed, only transformed. The changes in the energy of a system can be accounted for by taking the difference between the heat change in the system and the work done by the system.

3 Enthalpy (H): A more formal way to think about heat. The enthalpy (H) of a system its total energy, given by H = U + PV. This equation accounts for the energy needed to make the system plus the work needed to make room for the system in a constant-pressure environment. We can can t calculate absolute enthalpies, only changes in enthalpy -- H. For systems at constant pressure, in which no work occurs other than PV work (gas expansion), the enthalpy change in the system is equal to the heat change.

4 (+) and (-) signs in the first law: In an exothermic change, the system releases heat. The energy change has a (-) sign. H is negative. In an endothermic change, the system absorbs heat. The energy change has a (+) sign. H is positive. If the system performs work, like a compressed gas stretching a balloon, W, the work, has a (+) sign. If the system has work done on it, like a gas getting compressed inside a sealed syringe, W, the work, has a (-) sign.

5 3. Second Law of Thermodynamics: a) The entropy of the thermodynamic universe will always increase. b) The universe will always change in ways that increase the number of possible arrangements of the particles and energy. c) The universe will always change in ways that increase its randomness. Paraphrase: Even in a frictionless environment, heat cannot be converted to work with 100% efficiency.

6 The Heat of Neutralization Experiment: I) NaOH (s) NaOH (aq) H = (-32) kj/mole NaOH III) NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) H = (-87) kj/mole II) NaOH (s) + HCl (aq) H 2 O (l) + NaCl (aq) H = (-59) kj/mole If we add the heat changes from reactions 1 and 3, we obtain (roughly) the heat change for reaction 2. If we add the chemical equations for reactions 1 and 3, we obtain the chemical equation for reaction 2.

7 Hess s Law of Heat Summation (paraphrased): If we can add chemical equations to obtain a new chemical equation, then we can add the heat changes for these chemical reactions to find the heat change for the new reaction. I) A + B E + F H = X II) C + E A + F H = Y III) 2F + C B + G H = Z Sum: 2C G H = X + Y + Z

8 The law of heat summation allows us to calculate the heats of reactions without doing the reactions ourselves. 1. Define Standard heats of formation : The heat of reaction when a compound is made from its elements in their standard states. 2. The standard heat of formation of the elements in their standard states is zero.

9 Examples: Methane (CH 4 ): C (s) + 2H 2(g) CH 4(g) H f = kj/mole Water: H 2(g) O 2(g) H 2 O (l) H f = kj/mole Carbon Dioxide: C (s) + O 2(g) CO 2(g) H f = kj/mole

10 Now we ll find the heat of combustion of ethane without doing the reaction. Ethane is a component of natural gas. The balanced chemical equation: C 2 H 6(g) + 3.5O 2(g) 2CO 2(g) + 3H 2 O (l) H =?

11 Write the chemical equation as the sum of the equations of formation for the compounds, including the heats of reaction. We ll switch the signs of the heats when we write a formation equation in reverse. We ll multiply the heats by the balancing coefficients. C 2 H 6(g) 2C (s) + 3H 2(g) H f = kj/mole 2C (s) + 2O 2(g) 2CO 2(g) H f = 2(-393.5) kj/mole 3H 2(g) + 1.5O 2(g) 3H 2 O (l) H f = 3(-285.8) kj/mole

12 Now add the equations and add the heats, according to the law of heat summation. C 2 H 6(g) 2C (s) + 3H 2(g) H f = kj/mole 2C (s) + 2O 2(g) 2CO 2(g) H f = 2(-393.5) kj/mole 3H 2(g) + 1.5O 2(g) 3H 2 O (l) H f = 3(-285.8) kj/mole C 2 H 6(g) + 3.5O 2(g) 2CO 2(g) + 3H 2 O (l) H = kj/mole

13 The balanced equation for any* chemical reaction can be written as the sum of its equations of formation. Therefore, we can find the heat of any* reaction by summing the relevant heats of formation.

14 Thankfully, there s a big shortcut. C 2 H 6(g) + 3.5O 2(g) 2CO 2(g) + 3H 2 O (l) (-393.5) 3(-285.8) H reaction = Σ H products - Σ H reactants H reaction = [2(-393.5) + 3(-285.8)] - [-84.7] = kj/mole

15 Summary: We can add formation equations to obtain the equation for any* chemical reaction. Since we can add the formation equations to obtain the equation of interest, we also can add the corresponding heats of formation to find the heat of the reaction of interest. The short cut reduces the time that s needed to find the answer.

THERMODYNAMICS. Topic: 5 Gibbs free energy, concept, applications to spontaneous and non-spontaneous processes VERY SHORT ANSWER QUESTIONS

THERMODYNAMICS. Topic: 5 Gibbs free energy, concept, applications to spontaneous and non-spontaneous processes VERY SHORT ANSWER QUESTIONS THERMODYNAMICS Topic: 5 Gibbs free energy, concept, applications to spontaneous and non-spontaneous processes 1. What is Gibbs energy? VERY SHORT ANSWER QUESTIONS Gibbs energy (G): The amount of energy