Thermodynamics Ahmed Exothermic Rejoinders - Research Paper Example

Running header: Thermodynamics Student’s name: Instructor’s name: Subject code: Date of submission 1.The first law of thermodynamics The law concerns itself with changes in internal energy to heat added to a system as well as the work done by a system. The law is about conservation of enrgy and is represented by the equation below. First law of thermodynamics ΔU= Uf- Uo = Q- W In this case, U is the symbol representing internal energy, Q is positive when heat is added to the system and negative where heat is removed from the system, W is positive where work is done by the system and negative where work is done on the system. Work is force applied multiplied by the distance moved in the direction of force. Thus, the first law of thermodynamics states that a system’s total energy could neither be created or destroyed but it can be transferred from one location to another and even converted to and also from other forms of energy. The internal energy of the system is thus equal to the work being done on the system add or less the heat that flows in or out of the system and any other work being done on the system. 2.Entropy and its relationship to laws of thermodynamics Simply put, entropy referes to the measure of the amount of disorder in a sytem. It increases when the temperature of the system increases . For instance, when temprature increases thus meting an ice cube, the resultant arrangement of water molecules is more disordered and random as compared to when it was arranged in an ice cube (Yunus, 2014). Entropy is closely related to the three laws of thermodynamics. For instance, the second law and entropy are related in that the second law suggests that the world acts spontaneously to minimize potentials while maxmizing entropy . In all full natural processes, the entropy of the world always increases . 3.Explanation of the term spontaneous reaction The term spontaneous reaction refers to a reaction that happens without being triggered externally. In other words, it is a reaction which can occur under a given set of conditions but wothout application of external work. In the course of the reaction, heat is either absorbed from or released to the sorroundings without it requiring any outside assistance. For instance, Ice cubes will melt to water at +25 degrees centgrade . This is expressed as follows; H2O(s) H2O(l) is a spontaneous reaction at 250C and 1 atmosphere while H2O(l) H2O(s) is not a spontaneous reaction at 250C and 1 atmosphere At -25oC and 1 atmosphere however, the reverse of the above is true meaning that spontaneus reactions occur without external interference but only under a given set of conditions. 4. The terms endothermic and exothermic reactions The above terms differ in that endothermic reactions are the reactions which on occurring, the reactions absorb or take energy from the surroundings. This energy transfer is in the form of heat energy. As a result therefore, the reaction mixture as well as the surroundings gets colder. This causes the temperature to decrease and the decline can be detected by use of thermometer (Shankar, 2014). Such reactions include electrolysis, reactions between sodium carbonate and ethanoic acid and also when calcium carbonate is thermo decomposed in a blast furnace. On the other hand, exothermic reactions are opposite of endothermic reactions in that they are the reactions that when they occur, they transfer energy to the surroundings. This energy is similarly in the form of heat energy and it causes the reaction mixture as well as the surroundings to become warmer or hotter. Such reactions include neutralization reactions between alkalis and acids as well as the reaction between calcium oxide and water. 5. The three modes of heat transfer and how they differ from each other The three modes of heat transfer include conduction, convection and radiation The methods and how they differ are explained below; i) Conduction Conduction is the mode of heat transfer between two bodies (Mostly solid) that are in contact with each other such that heat flows from one body to the other. These bodies that conduct heat between each other are called conductors of heat. On the other hand, ii) Convection While in conduction heat itself moves hot portions of a fluid move through the fluid body in convection. The hot portions mix with the cold fluid and hence heat is transferred more quickly than is done in conduction. As a result of convection, hot fluids rise through surrounding, cooler fluids sink through warner fluids since they are denser (Bergman and Lavine, 2010). Thus, a circular motion of the fluid away from the heat source results. It is convection that drives ocean currents while also driving water patterns. Finally, iii) Radiation Radiation is considered the simplest mode of heat transfer in which case heat is carried by electromagnetic waves. Radiation is the movement of heat through a vacuum. 6. The difference in length of the two rods after heating (answer in mm) The difference in length between the two rods depend on a number of factors including the original temperature of individual rods and the temperature to which the individual rods are heated, the original length of individual rods and the coefficient of expansion of individual rods given that each rod expands differently (Tarik, 2010). In this case, Change in length = Coefficient of expansion *Original length *the change in temperature between the original temperature and the temperature to which the rods are heated. Thus; ΔL =α×L1×ΔT Where; ΔL = Change in length α = the rod’s coefficient of expansion L1 = Original length ΔT = Change in temperature Thus, the difference in length for the two metals will be given by the difference in the changes in length for the two metals = ΔL Aluminium – ΔL Iron = α×L1×ΔT Aluminium - α×L1×ΔT Iron = (23.1x10-6K-1× 1×161) – (11.8x10-6K-1×1×161) =0.000372 -0.00019 =0.000182 metres = 0.182mm The two rods will have a length difference of 0.182mm with Aluminium being the longer rod. 7. How many joules of heat energy are required to raise the temperature of 42.0g of gold from 23°C to 242°C? It is important to note that the heat energy requirement depends on the change in temperature to be achieved, the specific heat capacity of gold and the mass of water displaced as shown in the equation below; Q =mcΔT Where: Q represents Energy transferred in joules m represents mass of water displaced c represents the specific heat capacity of gold ΔT represents the change in temperature Thus; Q =mcΔT Q = 42 ×0.129×219 Q =1.1865× 103J Thus, 1.1865× 103J of heat will be transferred. 8. Calculate the specific heat capacity of mercury It is to be noted that the energy transferred in joules depends on the mass of water displaced, the specific heat capacity of mercury and the change in temperature when the mercury is immersed in the water as can be seen in the equation below; Q =mcΔT Thus, C (specific heat capacity) would be given by; C = Q/ (mcΔT) Where: c represents specific heat capacity of mercury Q represents the energy transferred in joules m represents mass of water and ΔT represents change in temperature. Thus; C = Q/ (mcΔT) c = 455/(25*130) c = 455/3,250 c = 0.14 c =0.14J/goC Thus, mercury has a specific heat capacity of 0.14J/goC 9. Calculate entropy ΔS, enthalpy ΔH and Gibbs free energy ΔG for the following equations: Entropy a) CaCO3(s) → CaO(s) + CO2(g) NB// each of the reactants and product has 1 mol. Therefore; ∆so = ∆so Cao(s) +∆so CO2(s)- ∆so CaCO3(s) = (39.8J/mol K + 213.7J/mol K)-92.9J/mol K = 253.5 J/mol K -92.9J/mol K =160J/K Enthalpy Cao(s) + CO2(g) → CaCO3(s) NB// each of the reactants and product has 1 mol. Therefore; ∆HfCaO(S) + ∆Sf CO2(g) - –∆SfCaCO3(s) =-635.09+ -393.50 --1206.9 =-1028.59+1206.9 =-178.31kj Gibbs free energy ΔG a) Cao(s) + CO2(g) → CaCO3(s) NB// Each of the reactants and product has 1 mol. Therefore; ΔG CaCO3(s) = 1ΔGf (CaO(s) + ΔGfCO2 (g) - ΔGfCaCO3 (s) = (603.5 +137.2)- -1128 = 740.7 --1128 = +387.5kj b) N2(g) + 3H2(g) ↔2NH3(g) Entropy NB//Since there are more than I mol of reactants and products; these are applied appropriately as follows; ∆so = 2mol∆so NH3 (g) - 1mol ∆so N2 (g) + 3 mol ∆so H2 (g) =2(192.78) - (191.5J/mol) + 3(130.6J/mol)) = 385.56-(191.5+391.8) =385.56-583.3 =-197.74J/K Enthalpy N2 (g) + 3H2 (g) ↔2NH3 (g) NB//Since there are more than I mol of reactants and products; these are applied appropriately as follows; ∆Hf N2(S) + 3∆Sf H2 (g) - – 2∆Sf (NH3(s)) =-(945+ -1308) - –2034 = -2253 --2034 =- 219kj Gibbs free energy ΔG NB//Since there are more than I mol of reactants and products; these are applied appropriately as follows; N2 (g) + 3H2 (g) ↔2NH3 (g) = ΔGf N2 (g) + 3ΔGf H2 (g) - 1ΔGf NH3 (g) =- 191.6+- 392.1-- 385.6 = -583.7+385.6 =- 198.1kj c) NH4NO3(s) → NH4+(aq) + NO3¯(aq) Entropy NB// each of the reactants and product has 1 mol. Therefore; ∆so = ∆so NH4NO3 (g) - ∆so NH4+ (aq) + ∆so NO3¯ (aq) =- 113.39J/mol+- 146.44J/mol --151.08Jmol = -259.83 +151.08 = -108.75 J/K Enthalpy NH4NO3(s) → NH4+(aq) + NO3¯(aq) NB// each of the reactants and product has 1 mol. Therefore; ∆HfNH4 +1(aq) + ∆Sf NO3-1 (aq) – (1∆Sf NH4NO3(s) = -132.51+-207.36--365.56 = -339.87+365.56 = 25.69kj Gibbs free energy ΔG NB/ Each of the reactants and product has 1 mol. Therefore; =1ΔGfNH4+1 (aq) + 1ΔGf (NO3-1 (aq)- 1ΔGfNH4NO3 (s) =-79.37+-111.34- -184.01 =-190.71-184.01 =6.7kj d) H2O(g) ↔ H2O(l) Entropy NB/ Each of the reactants and product has 1 mol. Therefore; ∆so = ∆so H20 (g) - ∆so H20 (l) = 188.7J/mol -69.9J/mol = 118.8 J/K Enthalpy H2O(g) ↔ H2O(l) NB/ Each of the reactants and product has 1 mol. Therefore; ∆HfH20 (g) + 1∆Sf (H2O (l) =685 -920 = -235kj Gibbs free energy ΔG NB/ Each of the reactants and product has 1 mol. Therefore; ΔGfH2O (g) - ΔGfH2O (l) =-228.6—237.1 = -8.5kj 10. How work and a change in energy are related Work and a change in energy are related in that work refers to an activity that involves a force and movement in the direction of the force. For instance, a force of 10 newtons pushing something 4 meters in the direction of the force will have done 40 joules of work (Turns, 2009). On the other hand, energy is the capacity to do work. As such, one has to have energy in order to do work. For instance, to do the 40 joules of work explained above, one would have to expend 40 joules of energy. In other words, for work to be done there must be a change in energy. Thus for every 1 joule of work done, it will imply a 1 joule change in energy. Explain how energy change and force are related. As stated above, work results when force moves an object in its direction for a certain distance. When work is done, some joules of energy will have been consumed in doing the work. Thus, force and energy change are related in that, force is used in doing work and when work has been done, it causes change in energy equivalent to the work done which in turn is a factor of the force expended to do the work. 11. Predict whether entropy increases or decrease for the following equations, include your reasoning. In solving this question, the following rules apply i) Entropy increases from solids to liquids to aqueous solutions to gases in that order. ii) As the number of moles of gases increase, entropy increases Thus, NaCl(s) → Na+(aq) + Cl¯(aq) - In this case, we move from solid to aqueous. Thus, entropy increases. 2NO (g) +O2 (g) → N2O4 (g) - In this case, we move from 3 moles of gas to 1 mole of gas. Thus, entropy decreases. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) - In this case, we move from gas reactants to a gas and liquid product. Thus, entropy decreases. 2NO2(g) ↔ N2O4(g) - In this case, we move from 2 moles of gas to 1 mole of gas.Therefore, entropy decreases. 12. Explain the Stefan-Boltzmann Law This law is used in relating the power that is radiated from a black body to its temperature. According to the law, the total energy radiated per unit surface area of a black body across all wavelength over a specific time length will be directly proportional the fourth power of the black body’s thermodynamic temperature. The law is given by the algebraic formula below; J = σT4 In this case, J represents the energy per unit area emitted by the body T is the body’s surface temperature in oK and σ is the Stefan- Boltzmann constant and is equal to 5.67 *10-8Wm-2K-4 What is emissivity? Emissivity refers to the ratio of energy that is radiated by the body to the energy that is radiated by a black body or a body that reflects no radiation with the same temperature. It is equal to absorptivity given that incoming energy is either reflected or absorbed. What is the range of values for the emissivity of a surface? Emissivity is zero for a perfectly reflective body and it is one for a black body. Thus, emissivity values for a surface range from 0 -1. Define the terms “black surface” and “grey surface”. A black body is a body or surface which is not reflective at all. In other words, a black body is a body with absorptivity of 1 and hence it absorbs the entire radiation without reflecting any (Howard, 2000). On the other hand, a grey body refers to a body that reflects some of the radiation and hence its absorptivity is less than 1. What role does the view factor play in determining the rate of heat transfer? View factor helps in expressing the fraction of radiation which leaves a certain surface and strikes another depending on its orientation subject to the other surface. What is a blackbody? Ideally, a blackbody is a physical body that is an ideal absorber and radiator of energy at all electromagnetic wavelengths. In other words, it absorbs all incident electromagnetic radiation without regard to frequency and angle of incidence. References: Yunus, C2014, Thermodynamics: An engineering approach, New York, McGraw Hill. Shankar, R2014, Fundamentals of physics: Mechanics, Relativity and Thermodynamics, Yale, Yale University Press. Bergman, T&, Lavine, A2010, Fundamentals of heat and mass transfer, London, Rutledge. Tarik, A2010, Engineering thermodynamics, New York, John Willey & Sons. Turns, S2009, An introduction to combustion: Concepts and applications, Western Australia, Cengage Learning. Howard, D2000, Thermodynamics and chemistry, Sydney, Prentice Hall. Read More