Calculate the equilibrium constant at 97°C for a reaction having ΔH = 17.39 kJ mol-1 and?
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!
249301
H3O+ .0292M = pH 1.53
874229
Heat released = 480*2.34 J\n\nSo 0.02 mol releases this much \n\n1 mol will release 480*2.34/0.02 = 56.16 kJ
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
400981
first, you find out the equation of Keq and delta G. I think it should be Keq = e^ [-deltaG/(RT)].\nThis equation and constant R can be found in any general chemistry textbook. And T is usu. 25+273=298K\nthen you can find out deltaG for the equation.\n\nwe also have deltaG= [- delta G for Ag(NH3)2^+1(aq) + (-32.0)] - [-26.2 - 3.8]\n\nand there comes the answer!!! ;)
122744
Mg + 2 CH3COOH = Mg(CH3COO)2 + H2
741784
(21 g x 52.7 C) + (54.9 g x 31.5 C) / (21 g + 54.9 g)\n\n= 37.4 C
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.
863200
Use the equation Q=mc(delta)T.\n36.8 = .006*130*(Tf - 32)\nSolve the equation for final Temperature and you get 79 degrees Celsius
250762
2 HCl + CaCO3 ----> CaCl2 + H2O + CO2
462663
C4H9OH + 6O2 --> 4CO2 + 5H2O
675016
This is a calorimetry problem! There is an equation for it:       \nq = mc(change in temperature)\n\nq = the heat required \nm= mass\nc= specific heat \n\nAlso, write out the reaction and products.
138065
The equilibrium constant of any equation is [C]^c[D]^d/[A]^a[B]^b.\n\nthe equation is already balanced. But you have to know how many moles of each element there are in order to compute the problem
136146
The 1st answer is wrong. The reaction is\n\nNa2CO3(aq) + SrCl2 (aq)--> 2NaCl (aq) +SrCO3 (s)
952215
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
739541
CH4 + 2O2-> CO2 + 2H2O
316742
Q=m.c.(delta)T\nQ=0.4 x 4200 x (20-16) = 6720 J\n\n6720=0.2 x c x (100-20) \nc= 420 J / kg K