Energy & Chemistry 2H2(g) + O2(g) → 2H2O(g) + heat and light
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Energy & Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 objects because of their difference in temperature. Other forms of energy — light electrical kinetic and potential Positive and negative particles (ions) attract one another. Two atoms can bond As the particles attract they have a lower potential energy NaCl — composed of Na+ and Cl- ions.
Energy & Chemistry 2H2(g) + O2(g) → 2H2O(g) + heat and light
This can be set up to provide ELECTRIC ENERGY in a fuel cell. Oxidation: 2 H2 → 4 H+ + 4 e- Reduction: 4 e- + O2 + 2 H2O → 4 OH- H2/O2 Fuel Cell. Energy, page 288.
HEAT is the form of energy that flows between 2 objects because of their difference in temperature. Other forms of energy — light. electrical. kinetic and potential. Positive and negative particles (ions) attract one another. Two atoms can bond. As the particles attract they have a lower potential energy. NaCl — composed of Na+ and Cl- ions.
Kinetic energy — energy of motion.
PE + KE = Internal energy (E or U) Internal Energy of a chemical system depends on. number of particles. type of particles. temperature. The higher the T the higher the internal energy. So, use changes in T (∆T) to monitor changes in E (∆E).
∆T measures energy transferred. SYSTEM. The object under study. SURROUNDINGS. Everything outside the system.
Heat always transfer from hotter object to cooler one. EXOthermic: heat transfers from SYSTEM to SURROUNDINGS. T(system) goes down. T(surr) goes up.
Heat always transfers from hotter object to cooler one. ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM. T(system) goes up. T (surr) goes down.
Energy & Chemistry. All of thermodynamics depends on the law of. CONSERVATION OF ENERGY. The total energy is unchanged in a chemical reaction. If PE of products is less than reactants, the difference must be released as KE. Energy Change in Chemical Processes. Potential Energy of system dropped. Kinetic energy increased. Therefore, you often feel a Temperature increase.
The heat required to raise an object’s T by 1 ˚C. Which has the larger heat capacity
How much energy is transferred due to Temperature difference The heat (q) lost or gained is related to. a) sample mass. b) change in T and. c) specific heat capacity. Specific heat capacity. = heat lost or gained by substance (J) (mass, g) (T change, K)
Substance. Phase. cp J g-1 K-1. Cp J mol-1 K-1. Air (typical room conditionsA) gas Aluminium. solid Argon Copper Diamond Ethanol. liquid Gold Graphite Helium Hydrogen Iron Lithium Mercury Nitrogen Neon Oxygen Uranium Water. gas (100 °C) liquid (25 °C) solid (0 °C) All measurements are at 25 °C unless noted. Notable minimums and maximums are shown in maroon text. Aluminum. A Assuming an altitude of 194 meters above mean sea level (the world–wide median altitude of human habitation), an indoor temperature of 23 °C, a dewpoint of 9 °C (40.85% relative humidity), and 760 mm–Hg sea level–corrected barometric pressure (molar water vapor content = 1.16%).
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al
q transferred = (sp. ht.)(mass)(∆T)
Drop into 225 g water at 21.0 ˚C. Water and metal come to 23.1 ˚C. What is the specific heat capacity of the metal
Note that T is constant as ice melts or water boils.
In general, reactions that transfer energy to their surroundings are product-favored. So, let us consider heat transfer in chemical processes.
heat energy transferred. ∆E = q + w. work done. by the. system. energy. change. Energy is conserved!
Exothermic reactions generate specific amounts of heat. This is because the potential energies of the products are lower than the potential energies of the reactants.
There are two basic ideas of importance for thermodynamic systems. Chemical systems tend toward a state of minimum potential energy. Chemical systems tend toward a state of maximum disorder. The first law is also known as the Law of Conservation of Energy. Energy is neither created nor destroyed in chemical reactions and physical changes.
(exothermic), -q. SYSTEM. ∆E = q + w. w transfer in. (+w) w transfer out. (-w)
Most chemical reactions occur at constant P, so. Heat transferred at constant P = qp. qp = ∆H where H = enthalpy. and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E. ∆H = change in heat content of the system. ∆H = Hfinal - Hinitial.
If Hfinal > Hinitial then ∆H is positive. Process is ENDOTHERMIC. If Hfinal < Hinitial then ∆H is negative. Process is EXOTHERMIC.
USING ENTHALPY. Consider the formation of water. H2(g) + 1/2 O2(g) → H2O(g) kJ. Exothermic reaction — heat is a product and ∆H = – kJ.
Liquid H2O. Making H2O from H2 involves two steps. H2(g) + 1/2 O2(g) → H2O(g) kJ. H2O(g) → H2O(l) + 44 kJ. H2(g) + 1/2 O2(g) → H2O(l) kJ. Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.
Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed. Active Figure
Hess’s Law of Heat Summation, Hrxn = H1 +H2 +H , states that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps. Hess’s Law is true because H is a state function. If we know the following Ho’s.
Notice that reaction [1] has FeO and O2 as reactants and Fe2O3 as a product. Arrange reactions [2] and [3] so that they also have FeO and O2 as reactants and Fe2O3 as a product. Each reaction can be doubled, tripled, or multiplied by half, etc. The Ho values are also doubled, tripled, etc. If a reaction is reversed the sign of the Ho is changed.
For any chemical reaction at standard conditions, the standard enthalpy change is the sum of the standard molar enthalpies of formation of the products (each multiplied by its coefficient in the balanced chemical equation) minus the corresponding sum for the reactants.
calculate Ho for the reaction below.
Hess’s Law Use a little algebra and Hess’s Law to get the appropriate Hovalues
Notice that the energy change in moving from the top to the bottom is independent of pathway but the work required may not be! Some examples of state functions are: T (temperature), P (pressure), V (volume), E (change in energy), H (change in enthalpy – the transfer of heat), and S (entropy) Examples of non-state functions are: n (moles), q (heat), w (work) ∆H along one path = ∆H along another path. This equation is valid because ∆H is a STATE FUNCTION. These depend only on the state of the system and not how it got there. V, T, P, energy — and your bank account! Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.
The properties of a system that depend only on the state of the system are called state functions. State functions are always written using capital letters. The value of a state function is independent of pathway. An analog to a state function is the energy required to climb a mountain taking two different paths. E1 = energy at the bottom of the mountain. E1 = mgh1. E2 = energy at the top of the mountain. E2 = mgh2. E = E2-E1 = mgh2 – mgh1 = mg(h)
Thermochemical standard state conditions. The thermochemical standard T = K. The thermochemical standard P = atm. Be careful not to confuse these values with STP. Thermochemical standard states of matter. For pure substances in their liquid or solid phase the standard state is the pure liquid or solid. For gases the standard state is the gas at 1.00 atm of pressure. For gaseous mixtures the partial pressure must be 1.00 atm. For aqueous solutions the standard state is 1.00 M concentration. ∆Hfo = standard molar enthalpy of formation. the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 6.2 and Appendix L.
2H2(g) + O2(g) → 2H2O(g) ∆H˚ = -484 kJ. H2O(g) → H2(g) + 1/2 O2(g) ∆H˚ = +242 kJ. H2(g) + 1/2 O2(g) → H2O(l) ∆H˚ = -286 kJ.
H2(g) + ½ O2(g) → H2O(g) ∆Hf˚ (H2O, g)= kJ/mol. C(s) + ½ O2(g) → CO(g) ∆Hf˚ of CO = kJ/mol. By definition, ∆Hfo = 0 for elements in their standard states. Use ∆H˚’s to calculate enthalpy change for. H2O(g) + C(graphite) → H2(g) + CO(g)
In general, when ALL enthalpies of formation are known, ∆Horxn = ∆Hfo (products) - ∆Hfo (reactants) Remember that ∆ always = final – initial.
Calculate the heat of combustion of methanol, i.e., ∆Horxn for. CH3OH(g) + 3/2 O2(g) → CO2(g) + 2 H2O(g) ∆Horxn = ∆Hfo (prod) - ∆Hfo (react)
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