Does anybody know how to find the enthalpy of an equation?
85318
Heat is equal to the change in enthalpy, dH, which is equal to the enthalpy of the products minus the enthalpy of the reactants.\n\ndH=H(products)-H(reactants)\n\n2B + 3F(2) --> 2BF(3)\n\nGet the heat of formation values for each specimen from a chemical thermodynamics table. I don't have a table in front of me so I'm just going to make up values so you'll know how to solve the problem. Let H=100 for B, H=150 for F(2), and H=400 for BF(3).\n\ndH= 2(400)-[2(100)+3(150)]\ndH=800-650=150 kj/mol\n\nI do know that elements alone have a heat of formation of 0, but I don't remember if the element has to be in a certain state or not.
709139
Bond enthalpy at 25 C\nC-H = 99 kcal/mole\nC-C = 83 kcal/mole\nC=C = 147 kcal/mole.\n\n(1st C) = 3 * 99 + 83 = 380 kcal/mole\n(2nd C) = 99 + 83 +147 = 329 kcal/mole\n(3rd C) = 2 * 99 + 147 = 345 kcal/mole
928379
(Triangle) or (Delta) H for this process will equal:\n\n60g * Cp of water (4.1855 J/(C*g) * (Delta T)\n\n60*4.1855*14* = 3515.82 J\n\nThe change in enthalpy or (triangle H) is 3515.82J
75392
You need to check your units.  The specific heat of water is not 4.184 J/g; it is 4.184 J/g*degrees C (or degrees kelvin).  And the heat of solution should not be in J/degree, but in J/g.\n\nTry this equation instead:\n\n[HoS = J/g] = [(Specific Heat = 4.184 J/g*C)*(mass = 100g)*(Temperature change = Initial C - Final C)] / [Mass of CaCl2 = 3 g]\n\nBy subtracting the final from initial you should get the negative value and not need the -1 at the front.\n\nSometimes enthalpy of solution is in J/mol rather than J/g, in which case you modify the above by calculating the mols of CaCl2 in 3 grams and switching that out with the 3g in the above equation:\n\n(3g) / (110.99 g/mol) = 0.027 mol\n\nTheoretically, you should want to know the heat capacity of the CaCl2 because it is also being heated, but this is assumed to be negligible.  Write that into your error analysis and you teacher will love you.  Also, make sure to check your significant figures; i don't know how many sig figs you used in weighing out your 3 grams, but that will be necessary to know how much to round everything.  Good luck.
615198
use the formula q=m*s*(t2-t1) where q is energy in heat,m is mass in grams,s is specific heat in joules/gram celsius,t1 is initial temp,t2 is final temp.\nsubstituting in formula,\n42,000=900*4.186*(t2-35)\nt2 works out to be 46.14 celsius.\nhope this helps.
390546
I'll give you a hint.\n\nDelta G = Delta H - T Delta S\n\nUsing that formula, you can get 1\n\n(B) you can look up in any physical chemistry text:\n\nG(2) - G(1) = a formula I don't have available
418645
heat of combustion = -(enthalpy change for reaction )\n          \n = -(heats of formation of products - reactants)\n \n1 mole of ethanol, when burnt completely, produces 2 moles of CO2 and 3H2O i.e delta H = 2(heat of formation of CO2) + 3(heat of formation of H2O ) \n\n= 2(-394 kJ/mol ) + 3(-286 kJ/mol)\n\n= -1646 kJ/mol\n\nheat of formation of ethanol = -277 kJ/mol\n\nso\n\nheat of combustion = - (-1646 - (-277) )\n                   \n                   = - (-1646 + 277 )\n\n                   = 1369 kJ mol
52130
E=mc∆T is the equation you want to use for this situation, where E is energy, m is mass, c is the specific heat, and ∆T is the change in temperature.  Now, the heat lost by the alloy is equal to the heat gained by the water, right?  So look what we can do:\nmc∆T=mc∆T\nwhere the left side of the equation is talking about the metal, and the right side of the equation is talking about the water.  Let's start pluggin in values.  We'll leave the left c value as a variable because that's what we're solving for.  \n36.0 g *c*(178 oC -30.6 oC)=185.0 g*(4.184 J/g*K)*(30.6 oC-25.0 oC)\nThe trickiest thing about that step is the temperatures: the metal started at 178 and goes down to 30.6; the water starts at 25 and goes up to 30.6.  \nNow evaluate that equation and solve for c. \nc=.817 J/g*K
958888
Use the formula:\nQ (heat) = Cp (Heat Capacity) * delta T (Change in Temperature)
434523
* The heat content of a chemical system is called the enthalpy (symbol: H)\n    * The enthalpy change (H) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure.\n    * H = H(products) - H(reactants)\n    * H is specified per mole of substance as in the balanced chemical equation for the reaction\n      The units are usually given as kJ mol-1 (kJ/mol) or sometimes as kcal mol-1 (kcal/mol)\n      1 calorie (1 cal) = 4.184 joules (4.184 J)\n    * Energy changes are measured under standard laboratory conditions\n      25oC (298K) & 101.3kPa (1 atmosphere) \n\nEnthalpy changes are measured in KJ per mole, but experiments on measuring enthalpy changes always measure a temperature change. The temperature change is then converted to enthalpy change using the following formula:\n\nDH    =    m x c x DT\n\nm = mass in grams\n\nc = specific heat capacity in Joules/Kelvin/gram\n\nDT = change in temperature in degree Kelvin\n\nThe experiment  is done by simple calorimeter experiments which involve carrying out the reaction in a heat resistant vessel and measuring the change in temperature. \n\nThis is also a common practical assessment, and you should be aware of the sources of error in your measurements such as heat loss to surroundings, incorrect readings, inaccurate thermometers etc.
234957
Check out this site: http://www.questconsult.com/~jrm/enthpres.html\n\nUse the component id for oxygen (28) and mole fraction (1.00), press submit form. Makes a nice Enthalpy-Pressure diagram.
809514
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
186407
One way to tell is to look up the heats of formation for the reactants and products.  Add all the reactants up and all the products up.  Subtract the reactants from the products.  If the answer is negative, the reaction is exothermic.  \n\nThe other way is to actually carry out the reaction and see if it liberates heat or not.
812302
You might just use the formula: \nEntropy of the reaction  = \n(sum of the entropy of products) - (sum of the entropy of reactants)\nIt's as simple as that :)\nThanks.
56221
The object of the game here is to make that final, last reaction out of the first three reactions (this will work because of a principle called Hess's Law, which states that enthalpy is a state reaction and will be the same, no matter how you get to that last equation). \n\nBecause you have to end up with four moles of GM as a product, take reaction (1) and double everything:\n2G2 + 2M2---> 4 GM  \nThat means that you have to double the heat, too, so the cooresponding enthalpy is 2(-851.0 kJ)=-1702 kJ. \n\nNow we also need two moles of LM4 as a product, so let's double reaction equation (2) to get the following:\n2L + 4M2--->2LM4 \nWe have to double the cooresponding enthalpy again: 2(-671.0 kJ=-1342 kJ.  \n\nFinally, we need to cancel out the L's and the G2's and add an L2G4 to the reactants of the final equation.  We will do this by flipping equation (3):\nL2G4--->2L + 2G2\nWhen we flip the equation direction, we also flip the sign on the enthalpy; the cooresponding enthalpy was 587.0 kJ originally, so now it's -587.0 kJ.  \n\nDo you see how all three of my new equations cancel out and add up to the final goal equation ? \n2G2 + 2M2-->4 GM  \n2L + 4M2--->2LM4 \nL2G4------->2L + 2G2\n____________________\nL2G4 +6M2-->2LM4 + 4GM\n\nNow, since we added up the equations to get the final equation, we can do the same thing with the enthalpies to get the final equation's enthalpy: \n-1702 kJ + -1342 kJ + -587.0 kJ = -3631 kJ!
1098900
The specific heat at constant volume for solid KCl as a function of temperature between 298 and 900 K is given by  (see 1st source):\n\nCp° (J/mol/K) = A + B*t + C*t^2 + D*t^3 + E/t^2\n\nwhere \n\nA = 35.41597 \nB = 70.03472 \nC = -91.38233 \nD = 52.52426 \nE = 0.153460 \nF = -449.1357 \nG = 108.9982 \nH = -436.6841\nand\nt = Temperature in K/1000\n\ndelta-H of formation of KCl at 298.15 = -436.68 kJ/mol (1st source)\n\ndelta-H of formation of KCl at 298.15 = -436.47 kJ/mol (2nd source\n\ndelta-H of formation of K+ (aq) at 298.15 = -252.17 kJ/mol (2nd source)\n\ndelta-H of formation of Cl- (aq) at 298.15 = -167.08 kJ/mol (2nd source)\n\ndelta-H of solution (using data from second source for KCl(s)):\n\n-252.17 +(-167.08) -(--436.47) = +17.22 kJ/mol\n\ni.e., this is the enthalpy change for the reaction:\n\nKCl (s) -> K+ (aq) + Cl- (aq)\n\nCan't help you with the value for Cv, but if you happen to know the molar volume, molar expansivity, and isothermal compressibility of solid KCl, at the temperature of interest, you can calculate Cv from the thermodynamic relationship:\n\nCp - Cv = T*V*a^2/K\n\nwhere T is the absolute temperature, V is the molar volume, a is the isobaric expansivity, and K is the isothermal compressibility.
804468
You need to look up the entropy for all compounds, multiply by the entropy by the stochiometric coefficients (e.g. 4 for NH3, 3 for O2). Then subtract the left side from the right side.\n\nVery simple actually, once you find the data.