What is the equilibrium constant (Ka) of NaHCO3?
599859
It can vary, based on the conditions. However, according to Harris's Quantitative Analysis (2002), it's 4.69*10^-11\n\nLet's be clear. That is the equlibrium constant for the reaction whereby the hydrogen carbonate loses a proton:\n\nHCO3- --> H+ + CO3-2
948592
Kc=2 = conc products / conc reactants\n\nKc = 2 = [CO][H2O]/[CO2][H2]\n\nStarting concs\n[CO] and [H2O] = 0\n\n[CO2] and [H2] = 0.7 moles/10L = 0.07M\n\nEquilibrium concs\n\nFrom the balanced equation, if x CO2 reacts, it reacts with x H2 to form x H2O and x CO\n\n[CO] and [H2O] = x\n\n[CO2] and [H2] = 0.7 moles/10L = 0.07M-x\n\n\nKc=2=(x)(x) / (0.07-x)(0.07-x)\n\nsolve for x and plug back into the final concentration equations.
425080
NaCO3(s) -->  2Na+(aq)  + CO3 2-(aq)
884595
For reaction aA <--> bB + cC, the Eq.constant K is defined as\n\nK = (B^b * C^c) / A^a\n\nSince in your equation a=b=c=1, and concentration for B=C, the equation simplifies to\n\nK = B^2 / A\n\nwith B=0.046 and A=0.454    (0.454=0.500-0.046)\n\nso K = 0.046^2 / 0.454 = 0.0047
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
942373
From the reaction\n\nS8(g) ===> 4S2(g)\n\nKp = (pS2(g))4/pS8(g) where p is the partial pressure\n\nGiven the partial pressure of S8(g) at equilbrium 0.25 atm, clearly 0.75 atm of S8(g) has been consumed to produce (from the equation) 4 volumes of S2(g), Given the volume was fixed this mean dS2(g) = 3 atm\n\nSubstituting into Kp yields\n\nKp =3^4/0.25 = 324\n\nThus Kp at 1325 K is 324
723847
I am assuming that the values of H° and G° are the standard enthalpy and Gibbs free energy changes for the formation of phsogene from CO(g) and Cl2(g).\n\nFor part (a), you should already know that:\n\ndelta-G = delta-H - T*delta-S\n\nSo, for the values you are given for 298 K,\n\n-206 kJ/mol = -220 kJ/mol - 298 K * delta-S°\n-46.98 J/(mol*K) = delta-S°\n\nFor part (b), we now have (assuming delta-H and delta-S are independent of temperature) that \n\ndelta-G_601K = -220kJ/mol - 601 K * (-4.698*10^-02 kJ/mol*K)\ndelta-G_601K = -191.8 kJ/mol
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
600936
Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O\n\nno. of mol of Na2CO3 reacted\n= m/M\n= 1.5/106.2\n= 0.01412 mol\n\nno. of mol of HCl reacted\n= 95/1000 x 6\n= 0.57 mol\n\nsince Na2CO3 is the limiting reagent, the no. of mol of CO2 liberated = 0.01412 mol\n\nMass of CO2 liberated\n= 0.01412 x 44.01\n= 0.6214g\n\nVol. of CO2 gas \n= 0.6214 x 1.96\n= 1.217\n= 1.22g (3s.f.)
288770
The Equilibrium Law\nIf the rates of the forward and backward reactions are equal at equilibrium, then it means that the concentrations of the reactants and products does not change.  The equilibrium constant Kc, is given by the product of the concentrations of the products (each raised to the power equal to that as the number of moles in the equation), divided by the product of the concentrations of the reactants (each raised to the power equal to that as the number of moles in the equation).  \n\nEg N2 (g)    +    3H2 (g)    2NH3 (g) \n\nKc =          The square brackets mean concentration in moldm-3\n\nor Kp =  The p is the partial pressure in atmospheres-1\n\nNB For equations involving a solids, the solid component is not included in the Kc expression.\n\nThis reaction is an example of a homogeneous equilibrium - all the reactants and products are the same state.\n\nKc is a constant at a given temperature and pressure.  If either of these factors are changed then the Kc value will also change.\n\n[click here for some worked examples]\ni am not in chemistry yet
958813
The KC value can be answered by knowing the equilibrium concentrations of NH3,N2 and H2.\nSince 2 mole of NH3 was taken initially and 1 mole of NH3 remains unreacted so moles of NH3 reacted :1.\nSince 1 mole of NH3 has reacted it means that 0.5 mole of N2 and 1.5 moles of Hydrogen has been formed in the reaction at equilibrium.\n\n              we have \n                            kc=[N2][H2]^3/[NH3]^2;\n                            [N2]=0.5;\n                            [H2]=1.5;\n                            [NH3]=1;\n                therefore\n                            kc=(0.5)*(1.5)^3/(1)^2;\n                 which gives Kc=1.6875\n                clarification to your doubt:\n                since the reaction taking place is:\n                    2NH3 <----->N2 + 3H2. \nthis means    NH3 <----->0.5N2 + 1.5H2. (divide the equation by 2 \n to get moles of N2& H2 produced per mole of NH3)\n   As 1 mole of NH3 reacted which means that 0.5 mole of N2 and   1.5 mole of NH3 is formed at equilibrium.Just substitute in Kc and find the  value as 1.6875.I think that answers your doubt.
119509
my question is:\n\nNaHCO3(s) + NaOH(aq) -> Na2CO3(aq) + H2O(l)\n\n3.25 moles of Na2CO3 is produces.
158900
Yup. They are not simple...\nChemistry...\nhttp://www.chm.davidson.edu/ChemistryApplets/equilibria/EquilibriumConstant.html\nhttp://en.wikipedia.org/wiki/Chemical_equilibrium\nThey use the equilibrium constant expression (Kc) here...\nhttp://www.cofc.edu/genchemlab/equilibrium.htm
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!!! ;)
543314
1.) Q = (nh3)^2/(n2*h2^3) = 0.16\n\nAs Q>Kp, reaction will move backward.\n\n2.) Kc = Kp*(RT)^(1+3-2) = 3.1*10^-4*(8.314*700)^2 = 3.55*10^11\n\n3.)\n\nGiven reaction is 2*(ii)+(i)\n\nSo, Kp = (8.3*10^-3)^2*3.1*10^-4=2.13*10^-8\n\nLet original P of HCN = 1\n\nso, partial pressure of C2N2 and H2 = p (assuming 2p of HCN decomposes at equilibrium)\n\nEqn is\n\np*p/(1-2p)^2 = 4*10^-4\n\nSo, p/(1-2p) = 2*10^-2\n\nSo, p = 0.019 atm.\n\nEnjoy.............
304632
Okay, so the first thing that happens it that the NaF completely dissociates:\nNaF ---> Na+ + F-\n\nCool.  Now, the Na+ is just floating around, so we can ignore it.  Let's deal with the F-.  The F- is in the water, so it's going to do this:\nF- + H2O <---> HF + OH-\nNote that the arrow is going in both directions; it's an equilibrium problem, so I need to set up an equilibrium equation.  \n\nBut first, because I'm dealing with OH- ions, I'll need a Kb value.  Since Kw, the product constant of water, equals Kb * Ka, and Kw= 1E-14:\n(1E-14)/(6.8E-4)= Kb = 1.47E-11\n\nNow,\nKb=[OH-][HF]/[F-]\nor\n1.47E-11 = x^2/(.00339-x)\nsolve for x (I just set it equal to zero, graph it, and find the x-intercpets on my calculator):\nx=2.23E-7 M \nSince x=[OH-], [OH-]=2.23E-7 M\n\nWe can calculate pOH by taking the negative log of [OH-]:\n-log 2.23E-7 M = 6.65\nSince pH + pOH = 14.00:\npH + 6.65 = 14.00\npH = 14.00-6.65\npH = 7.35, the answer within one one-hundredth, which is probably some minor rounding discrepancy.  Yay!