Can anyone give information on how to solve Thermochemistry problems?
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In thermodynamics and physical chemistry, thermochemistry is the study of the heat evolved or absorbed in chemical reactions. Thermochemistry, generally, is concerned with the heat exchange accompanying transformations, such as mixing, phase transitions, chemical reactions, etc., which includes calculations of such quantities as the heat capacity, heat of combustion, heat of formation, etc. The laws of thermochemistry rest on two statements:[1]\n\n   1. Lavoisier and Laplace’s law (1780): the heat exchange accompanying a transformation is equal and opposite to the heat exchange accompanying the reverse transformation.\n   2. Hess’s law (1840): the heat exchange accompanying a transformation is the same whether the process occurs in one or several steps\n\nBoth laws preceded the first law of thermodynamics (1850); it can be shown, however, that they are a direct consequence of it.
518074
I have only problems can you help me to solve some ?
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http://www.chemtutor.com/\nthis was a site that helps all in the basics That is how i learned it.
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
315436
i do not think that there is a solution book for McKelvey. I have not seen any. u can try amazon and check though but that will not be free (if it exists). i would think that some of the problems in the QM outline chapter will be common to that of Merzbecker (some may be solved). For the semiconductor part try looking in Sze, Ashcroft and Mermin  and check if the solved problems have them.
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Here is a good guide\n\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Limiting-Reagent.html
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Try these links:\n\nhttp://chemweb.calpoly.edu/chem/124/124Experiments/Thermochem/HessExp.html\n\nhttp://services.juniata.edu/ScienceInMotion/chem/labs/ap/hess.doc
278691
Use the combined law of gases formula P1*v1/T1=P2*v2/T2.  Where P=pressure, v=volume, and T=temperature.  If the problem doens't have either the temperatures or the pressures assume them to be 1.  I.E. The answer to problem number 7 is:  8 liters.
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rubbing 2 things together to create heat.\nRub yor hands together really fast and see what happens.
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Your answer depends on the nature of your problem, so you'll want to give more info if you want a complete answer.  The specific heat capacity you use depends on which substance is being heated up or cooled off.  The mass will be the mass of the substance that is being heated up or cooled off.\n\nOne exception that frequently occurs is when you are dissolving a substance in water.  Most problems will assume that the solution has the density and specific heat capacity of pure water.  So in other words, if you dissolve 3 grams of ammonium chloride in 100 mL of water and watch the temperature go down, most instructors would want you to use a mass of 100 grams (instead of 103) and the specific heat capacity of water 1.00 cal/g C or 4.184 J/g C if you're working in joules.\n\nHope this helps...this is the best I can do without more info about the problem.
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Just let the units tell you what to do.  H=mass* T*sp heat.  with T in K
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You might try one of "Shaum's Outline Series" on thermodynamics or physics etc. Here's a list of titles:\n\nhttp://btobsearch.barnesandnoble.com/booksearch/isbnInquiry.asp?z=y&btob=Y&isbn=0070507074&itm=1
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Problem \ntake D as Delta(change)\nEstimate the change in enthalpy, DH, for the following reaction:\n\nH2(g) + Cl2(g) --> 2 HCl(g) \n\nSolution \n\nTo work this problem, think of the reaction in terms of simple steps:\n\n\nFirst, the reactant molecules, H2 and Cl2, break down into their atoms\nH2(g) --> 2 H(g)\nCl2(g) --> 2 Cl(g)\n\n\nNext, these atoms combine to form HCl molecules\n2 H(g) + 2 Cl(g) --> 2 HCl(g) \n\nIn the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction:\n\nDH1 = +(436 kJ + 243 kJ) = +679 kJ\n\nBond breaking requires energy, so we expect the value for DH to be positive for this step.\n\nIn the second step of the reaction, two moles are H-Cl bonds are formed. Bond breaking liberates energy, so we expect the DH for this portion of the reaction to have a negative value. Using the table, the single bond energy for one mole of H-Cl bonds is found to be 431 kJ:\n\nDH2 = -2(431 kJ) = -862 kJ\n\nBy applying Hess's Law, DH = DH1 + DH2\n\nDH = +679 kJ - 862 kJ\n\nDH = -183 kJ\n\n\nAnswer\n\nDH = -183 kJ
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I got my Ch.E. this last May and I remember having trouble with Heat and Mass Transfer.  I was lucky enough to have a REALLY good professor who could teach anything to anyone!  The book we used was Incropera & Dewitt "Fundementals of Heat and Mass Transfer".  I thought it was a pretty good book.  You may be able to find it around.\n\nYou could also try getting help from upperclassmen.  At the school I went to (Minnesota Duluth) there was a lot of cooperation between students.\n\nI don't think there are any websites that would clearly and thoroughly explain this subject to you.  An entire semester ins't even enough to fully understand it!
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Here you are:\n\nFirst write the combustion equation and balance it:\n\nH3NNH2(l) + O2(g) ---> N2(g) + 2H2O(g).\nThe heat will be the diffrence of enthalpies of formation between the products and the reactants. Do not forget to multiply by the stoichiometric coefficient.\n\nRemember that the enthalpies of formation of O2 and N2 are 0.\n\nYou will also need to find the enthalpy of formation of H2O(g) in the table of your book.  It is -241.8 kJ/mol\n\nHf(H2NNH2) = [2*(-241.8) + 1*0] - [1*50.6 + 1*0]\n\nHf(H2NNH2) = -483.6 - 50.6 = - 534.2 kJ/mol
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1- find heat quantity\n Q=C*t\nQ = heat quantity of the reaction\nC= heat capacipy of calorimeter\nt=temperature increasing\n\n2- find heat of reaction \n H=(Q*1000)/(N*V)\nN = normality of acid and base\nV= volume of acid and base
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http://www.sbuniv.edu/~ggray/CHE3345/chp14.html\nhttp://web.odu.edu/sci/tisenhou/chem321/chap12/index.htm\n\nbest luck