What is the general equation for finding ΔH of a reaction from ΔHf°?
813100
ΔH =ΣΔHf°prod-ΣΔHf°react
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.
905994
deltaG = deltaH - TdeltaS\n\nWhen delta G equals zero,\n\ndeltaH = TdeltaS\n\ndeltaS = deltaH/T  = (44.0 kJ/mol)/373K = 0.11796 kJ/mol.K\n\nSo I'd pick e.
835897
Assuming KOH is a strong base,\npOH = - log [-OH] = - log (1.00 M) = 0\npH + pOH = 14\npH = 14 - pOH\npH = 14\npH = - log [H+]\n[H+] = antilog(-pH)\n[H+] = 1 x 10^-14 M
565612
Take all the terms that include H to one side.\n\nS - 2LW = 2LH + 2WH.\n\nNow, factor out the H from the right side to get:\n\nS - 2LW = (2L + 2W)*H.\n\nNow, divide both sides by what it takes to leave H by itself.  In this case, get rid of the (2L + 2W)\n\n(S - 2LW)/(2L + 2W) = H
757438
CaC2 + 2 H2O ---> Ca(OH)2 + HCCH
724209
Delta Hf = Delta Hf of the products - Delta Hf of the reactants\n\nThe Delta H's of pure species in their natural form are zero. That means that O2 and H2 will have delta Hf's of zero.\n\nThe other values you have to look up.\n\nOverall it will be...\n\nDelta Hf = 2*Delta H of CO - 2*Delta H of CH4\n\nThe twos are the stoichometric coefficients.
1118780
Your question is very open.\n\nheat equations very general can be:\napplied heat = specific heat x mass x increase of temperature\napplied heat = heat capacity x increase of temperature\napplied heat = mass x specific melt heat\napplied heat = mass x specific evaporation heat (at boiling point)\nHeat from a supply = power x time.\nCombinations and others are possible.\n\nwave equation can be x = A sin( 2pi t/T)\nThere are more possibilities.
970544
Here is what you do: \nDo the same operations as I did on the equations with the H values and you will be OK (Do not forget to use the same multipliers on the H as I did on the reactions).\n\n10PCl3 + 5O2 ---> 10Cl3PO\n10PCl2 + 10Cl2 --- 10PCl5\nSubstract the equations to get (keep coeff positives by moving to the other side):\n\n5O2 + 10PCl5 ---> 10Cl3PO + 10Cl2\nP4 + 5O2 ---> P4O10\nSubstract and get\n10PCl5 + P4O10 ---> P4 + 10Cl3PO + 10Cl2\nP4 + 6Cl2 ---> 4PCl3\n\nSum the two to get (P4 goes):\n\n10PCl5 + P4O10 ---> 10Cl3PO + 4Cl2 + 4PCl3\n\n4 PCl3 + 4Cl2 ---> 4PCl5\n\nSum the two equations\n\n10PCl5 + 6P4O10 ---> 10 Cl3PO.
351293
it is based on arrhenius equation\nK=A e ^ (-Ea/RT)\nA- frequency factor\nEa- activation energy (proportional to Delta H)\nR -Gas constant\nT -Temperature\nSince you dont know A- fequncy fator, yu need to do a lengthy calculation\n\nThe modified arrhenium equation will be ( by considering two situations T1, T2 there will be two rates k1, k2 but A-frequency factor is smae for one reaction so it's a constant..)\n\nEa= (R X T1 X T2 X Ln (k1/k2) ) / (T1-T2)\n\nEa= delta H = 17,390 J\nR = 8.314 J/K\nK1 = 1.6 X 10E-4\nK2= ?\nT1 = 298\nT2= 273+97= 370K\n  After substituing all these values, you will get K2= 6.27 E-4\nwhich is in agreement with theory (At high Temperatures, rate should be high..!)\n\nIs this info will help you?\nGood luck!
350981
This isn't too bad.\n\nFirst, find the enthalpy of formation for each of the reactants and products.\n\nFe2O3(cr) = -272 kJ/mol\nCO(g) = -110.525 kJ/mol\nCO2(g) = -393.509 kJ/mol\nFe = 0 kJ/mol\n\nThen add the enthalpies of formation for the reactants and subtract the enthalpies of formation for the products:\n\nFe2O3 + 3CO - (3CO2 + 2Fe)\n\n-272 kJ/mol + 3(-110.525 kJ/mol) - 3(-393.509 kJ/mol) - 2(0) = -603.575 kJ/mol - -1180.53 = -576.952 kJ/mol \n\nThat should do it.
868428
ΔP.E. = Δmgh\nwith m and g constant,\nΔP.E. = mgΔh\n\n(a)\nE(A) = 50*10*9.8 = 4900 J\nE(B) = 0\nΔE = 4900 J\n\n(b) and (c) have similar solutions.
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!
747676
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4
887797
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
752218
2KOH + Cl2 ----> KCl + KClO + H2O\n6KOH + 3Cl2 ----> 5KCl + KClO3 + 3H2O