What is the enthalpy for the reaction?
921703
Just let the units tell you what to do.  H=mass* T*sp heat.  with T in K
871220
The second one:\nAF + H --> AFM + TM\n\nActivation E: +10KJ/mol is moderate\nHeat of Rx= +5.2 KJ/mol is positive, making this an endothermic reaction (requires the input of heat or energy to accomplish).\n\nSee the site below:
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!
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.
288242
The reaction should be written as follows,\n2 KClO3(s) --> 2 KCl(s) + 3 O2(g) H = -89.4 kJ\n\nYou either did not include the arrow or it is not displaying properly.\n\nPotassium Chlorate decomposes with heat to form Potassium Chloride and Oxygen gas, no where in this reaction is elemental Potassium involved.\n\nThe standard enthalpy of formation of any pure element in its natural state is zero. Oxygen is a diatomic molecule and thus, Oxygen gas (O2) has a enthalpy of formation of 0 kJ in this reaction, no matter how much of it you have, it is still zero.\n\nSince the Potassium Chloride is the only product left that has any non-zero enthalpy of formation, we know that the over all change in enthalpy is equal to the enthalpy of formation of KCl * the number of moles produced.\nIf the over all change in enthalpy is -89.4 kJ and 2 moles of KCl are produced, then the enthapy of formation of the KCl must be 44.7 kJ/mole less than the enthalpy of formation of the Potassium Chlorate.
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
607425
Yes, provided the conditions are right. For example if the reaction\n\nA + B -> c + d\n\nis exothermic, the reaction in reverse will be endothermic, i.e. you will have to supply the heat.
136146
The 1st answer is wrong. The reaction is\n\nNa2CO3(aq) + SrCl2 (aq)--> 2NaCl (aq) +SrCO3 (s)
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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 )
278240
C6H12O2 + 6O2 --> 6CO2 + 6H2O\n\nis the combustion reaction for glucose.
294164
Thw formular to use here is Q = mcT where Q = quantity of heat, m = mass, c = specific heat capacity, and T is the change in the temperature, thus here Q = 2.0g*(30-25)*0.90 = 9J. Notice that I sed Joules as the unit.\n\n2.For this one, a law which i cant remember the name right now is nedeed. Here we have C(s)+O2(g)=CO2 Enthalphy -390KJ\nand also Mn(s)+O2(g)=MnO2(s) Enthalphy = -520KJ\nFor this law to be appplied, we have to write the second equation the other way to become MnO2= Mn+O2 Enthalphy = 520KJ. Notice that i changed the sign when i reversed the equation. Then O2 could be canceled out on the left side of the first equation and on the right side of the second equation. This was we are adding. Finally adding, we get MnO2(s)+C(s) = Mn + CO2 Enthalphy = -390KJ+520KJ = 130KJ.\nand thats the correct answer
863456
I assume the reaction in question is:\n\nA(g) <-> 2B(g)\n\nThe equilibrium constant, Keq, for this reaction is given by:\n\nKeq = ([B]^2)/[A]\n\nwhere the quantities in square brackets mean "the concentration of" the relevant species at equilibrium (or, in the case of gases, one commonly uses equilibrium partial pressures instead of concentrations).\n\nPlugging in the values we are given for the equilibrium concentrations, we have that:\n\nKeq = (0.154^2)/0.381\n\nKeq = 0.0622
871634
stndrd enthalpy of formation for NaF (s) is -569.0 kJ/mol\n\nhttp://chemistry.about.com/library/weekly/blheatform.htm
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reactant(s)\nSynthetic
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H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
908972
Any combustion reaction is exothermic.\nSo a and b are exothermic.\nc is the dissociation of water, which involves breaking bonds, so it is endothermic.\nI think d is supposed to represent the melting of ice, which again takes in heat.
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.