What is the difference between the molar heat of reaction and heat of reaction?
82812
heat of reaction - total heat produced or used during a reaction\nmolar heat - heat produced or used during a reaction PER MOL
309732
i.e. it is easier to go backwards than forwards, so the product has more energy ( i.e. higher up the energy scale / nearer the top/ easier to get to the transition state ).\n\nHeat of reaction = heats of formation of products - reactants = 85 - 80 = +5kJ mol.... endothermic ( makes sense if products contain more energy than reactants )
963248
The heat given off by the burning reaction (delta E) is equal to the heat gained by the calorimeter.  (m * change in temp)\n\nYou can calculate the heat released by burning.\n.553 g with a molecular wt. of 88g/mole \n(.553 grams) / (88 (grams / mole)) = 0.00628409091 moles\n(975 (kJ / mole)) * 0.00628409091 moles = 6 126.98864 joules\n\nYou can calculate the heat absorbed by the water (3000 g * .356)\n3 000 * .356 = 1 068 joules\n\nyou can calculate the difference, it must have heated up the apparatus.\n6 127 - 1 068 = 5 059 joules \n\n5059 joules = x grams * .356\n5059 joules / .356 = 14210.67 grams of water\n\nThe metal apparatus was had a heat capacity equivalent to 14210.67 grams of water.
816266
At constant temperature and pressure, a reaction will occur spontaneously if the Gibbs free energy change of the reaction is less than zero.\n\nThe Gibbs free energy change, delta-G, is given by:\n\ndelta-G = delta-H - T*delta-S\n\nwhere T is the absolute temperature, \n\ndelta-H is the enthalpy change for the reaction at the temperature and pressure of interest, and \n\ndelta-S is the entropy change for the reaction at the temperature and pressure of interest.\n\nWe are given the enthalpy and entropy changes for the reaction of interest in this question.  I will assume that the enthalpy change and entropy change for this reaction are independent of temperature. Note also that the values of both of these thermodynamic parameters are given in "per mole" units in this question.  I will assume these values are normalized to one mole of one of the reactants, and not one mole of the product.\n\nPlugging in the enthalpy and entropy values, and setting delta-G to be less than zero:\n\n0 > delta-G = 180500 J/mol - T*24.8 J/(mol*K)\n\nT  > (180500 J/mol)/(24.8 J/(mol*K))\n\nT > 7278 K\n\nThis is an extremely high temperature, and the assumption that we made above that the enthalpy and entropy changes were independent of temperature is probably not very good given the large temperature extrapolation being made here.
341402
Specific heat does not apply to reactions, only actual chemicals.  You are probably thinking of the heat of reaction, or enthalpy.  The change in enthalpy of a reaction (written as delta H, delta as the Greek letter that looks like a little triangle) flips its sign when the reaction is run in the reverse direction.  So a positive delta H value becomes negative when the reaction is run in reverse direction, and a negative delta H value becomes positive.
952146
The first part of the question gives you the information you need to determine the "calorimeter constant", or the amount of heat required to change the temperature of the whole calorimeter set-up by 1 degree.\n\nThe calcium carbonate/HCl stuff is the reaction whose enthalpy change you are trying to determine, using the same calorimeter as before.  When you combine calcium carbonate and HCl, you'll get a reaction that produces CO2 and leaves you with a solution of calcium chloride.  So the overall reaction that you're looking at is:\n\nCaCO3 + 2 HCl --> CO2(g) + H2O + CaCl2\n\nCalcium carbonate will only decompose to CO2 and CaO when you heat it--not when you react it with an acid.\n\nHope this helps....I'll let you do the calculations.
984673
molar mass Fe=55.85\nmoles iron=490/55.85\nmoles iron=8.774\n4 moles are required for the reaction\n8.774/4 provides the number of times the reaction can be completed\nthis equals 2.19\nmultiply this by -1.65103\nand you get -3.62 kJ.  This being negative tells you the reaction is exothermic and that the positive 3.62kJ is the amount given off
747557
A reaction tends to be spontaneous if  it causes an increase in the entropy of the system, and/or a decrease in the enthalpy of the system \n\nBoth reactions are endothermic (increase in enthalpy) becuase of positive delta H values.\nBoth have an increase in entropy (positive delta S value for the first one, production of a gas for the second)\n\nSo I'd say statements ll and lll are true.
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
706503
find the heat evolved, assuming heat cap of water is 4.2J/g/K\n\n(31.43-25.32) x 4.2 x 100g = 2566.2J\n\nnow according to definition, molar heat of neutralisation is calculated upon the mols of water formed. this number of mols is equivalent to the no. of mols of the limiting reagent (in this case, the HCl).\n\nmol of HCl= 0.050 x 1 = 0.05mol\n\nenthalpy of neutralisation= 2566.2/0.05= 51324J/mol\n\nthis answer is pretty close to real-life experimental values for strong acid-strong base.
1138895
They are similar, but not exactly the same as exothermic and endothermic as used in chemistry.\n\nExergonic reactions are defined by a negative delta G, that is they have less free energy at the completion of the reaction than at the beginning. An exergonic reaction may require some input of energy (called activation energy) to start the reaction, but will release more energy than is required to start the reaction. \n\nAn endergonic reaction will have a positive delta G, that is it will have more free energy after the reaction than before it starts. \n\nHere is a good website with graphs that explains this:\nhttp://academic.brooklyn.cuny.edu/biology/bio4fv/page/exergon.htm\n\nHere is another website that explains the similarity between exergonic, exothermic, and endergonic, endothermic. \nhttp://www.mansfield.ohio-state.edu/~sabedon/sta5202.htm
676662
What did you use to measure the activation energy?\nYou probably used a thermometer which was probably marked to 0.1 degrees. That means that your measurements are probably accurate to 0.05 degrees. Reactant temp may be measured as 25.1 (+/- 0.05) so this means that the temp could be 25.05 or 25.15.\nDo the same for the transition state.\nDraw small vertical bars through each point, e.g for reactants the bar would go from 25.05 to 25.15, with the cross in the middle.\n\nI can't remember the formula, but errors are a lot more pronounced on small measurements. It may be something like the difference between the highest and lowest possible values divided by the recorded value, all multiplied by 100.
444111
Rate of reaction depends upon the concentration of reactants. Since, 6M HCl has a greater conc. than 3M HCl, rate of reaction decreases when Mg reacts with 3M HCl
596796
Exothermic reactions produce heat, so the temperature of the reactants will become higher, i.e. hotter.\n\nEndothermic reactions absorb heat from the surroundings, so the temperature of the reactants will become lower, i.e. colder.\n\nFor a specific reaction involving Co(II), would have to know the reaction equation and then calculate the delta enthalpy.
92261
1.  Heat speeds up a reaction by increasing vibrations and energy of the reactants.\n\n2.  Concentration speeds us a reaction by making more reactants available in a unit volume.\n\n3.  Catylists speed up a reaction (without being used up) by allowing reactants to come into closer proximity.\n\n4.  Increased surface area (by grinding, etc.) speeds up reactions by making more reactants exposed to reaction.
533621
(delta)H(rxn) = sum of heat of formation of products - sum of heat of formation of reactants.\n\na.  Hf Br2(g) = 30.91 kJ/mol, Hf Br2(l) = 0, so (delta)H = -30.91 kJ/mol\n\nb.  Hf CaCO3(s) = -1207 kJ/mol, Hf CaO(s) = -635.5 kJ/mol, Hf CO2(g) = -393.5 kJ/mol, so (delta)H = -635.5 - 393.5 + 1207 = +178 kJ/mol\n\nc.  Hf NO(g) = 90.25 kJ/mol, Hf O2(g) = 0, Hf NO2(g) = 33.2 kJ/mol, so (delta)H = 2 * 33.2 - 2 * 90.25 = -114.1 kJ/mol
84780
!!!!!!!!!!!!!!!! \nI think rather u wanted to ask how can u predict the formula of a chemical reaction!!\nFirstly, we have to determine the moles of reactant taking part.(It can be done easily, think A reacts with B, just mix 1 mole of A with the same amount of B.)\nThen determine what are the different products and their amounts.\nAll u have to do is some calculations..................\nRemember its not as easy as it sounds.