Equilibrium question?
922031
The reaction for the Fritz-Haber-Bosch process is:\n\n3H2(g) + N2(g) <-> 2NH3(g)\n\nThe equilibrium constant for this reaction can be written as:\n\nKp = ((p_NH3)^2)/((p_N2)*(p_H2)^3) = 7.16*10^-5 atm^-2,\n\nwhere p_xxx stand for the partial pressure of the gas "xxx" at equilibrium.\n\nWe are told that initially (before any reaction) the partial pressures of H2 and N2 gases are 3.7 atm and 5.4 atm, respectively.  Because each mole of NH3 produced consumes 1/2 mole of N2 and 3/2 mole of H2, we know from mass balance that at equilibrium,\n\np_N2 = 5.4 atm - 0.5*p_NH3\n\nand\n\np_H2 = 3.7 atm - 1.5*p_NH3\n\nPlugging these expressions into the equilibrium constant, we have that:\n\n7.16*10^-5 atm^-2 = ((p_NH3)^2)/[(5.4 atm - 0.5*p_NH3)*(3.7 atm - 1.5*p_NH3)^3]\n\n\nlet p_NH3 = x, so the algebra is easier to write out.\n\n7.16*10^-5 atm^-2 = (x^2)/(273.53 atm^4 - x*357.99 atm^3 + x^2*165.67 atm^2 - x^3*30.712atm + x^4*1.688)\n\n\n(7.16*10^-5 atm^-2)/x^2 = 1/(273.53 atm^4 - x*357.99 atm^3 + x^2*165.67 atm^2 - x^3*30.712atm + x^4*1.688)\n\nx^2/(7.16*10^-5 atm^-2) = 273.53 atm^4 - x*357.99 atm^3 + x^2*165.67 atm^2 - x^3*30.712atm + x^4*1.688\n\nx^2*1.397^4 atm^2 = 273.53 atm^4 - x*357.99 atm^3 + x^2*165.67 atm^2 - x^3*30.712atm + x^4*1.688\n\n273.53 atm^4 - x*357.99 atm^3 - x^2*1.38*10^4 atm^2 - x^3*30.712atm + x^4*1.688 = 0\n\nThis is an ugly 4th order polynomial in x.  The most direcet way of solving for the roots is to simply plot it, and look for the zeros of the function.  I'll admit to using a numerical solver to find that the roots are at:\n\nx = -81.762 atm, -0.1548 atm 0.1284 atm, and 99.98 atm.\n\nOnly the positive roots have a physical meaning, and the root at 99.98 atm is not possible because the starting pressures of the reactants are not high enough to generate a partial pressure of the product that is this high.  We have, then, that x = p_NH3 = 0.1284 atm.\n\nWe can check that this is a correct answer by calculating the equilibrium partial pressures of N2 and H2 and plugging all the values back into the equilibrium constant:\n\nK = ((0.1284 atm)^2)/((5.4 atm - 0.5*0.1284 atm)*(3.7 atm - 1.5*0.1284 atm)^3)\n\nK = 7.16*10^-5 Check!\n\n--------------\n\nThe previous answerer had the right idea, but has the wrong expression for the equilbrium constant because (s)he wrote the chemical reaction so that one mole of NH3 is produced (i.e., the reaction I wrote divided by two).  The equilbrium constant for that reaction would have units of atm^-1, not atm^-2, as we are given in the question.  Had the previous answerer used the square root of the equilibrium constant given in the question, then his/her answer would have been correct.
877878
yes. This causes an imbalance in the equilibrium, so the equilibrium will shift to the right to compensate for this. The reactants would try to make more products to compensate for the product being taken away, so this would be an example of the equilibrium shifting, which is le chatlier's principle =)
683249
Oh, come on, it doesn't get any easier than this!\n\nSnO2 + 2H2 --> Sn + 2H2O\n\nAbove is balanced. 1 Sn on both sides, 2 O on both sides and 4 H on both sides.\nYou need to work a little harder on these yourself.
785530
Answer D.\n\nThis is the Nernst equation at standard conditions.
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::
378344
At equilibrium, we have this below relationship between K and the concentration of all species.\n     \n　　　　　　　　　　　　　[H2]^2 [O2]\n　　　　　　　　　　　K= ----------------\n　　　　　　　　　　　　　　[H2O]^2\n\nSo we could get the answer by solving this equation.
660890
If it is balanced, it is in equilibrium by definition.
447650
Equilibrium is defined as the state of undergoing equal forces in all directions (that is mathematically equivalent to zero force).  A moving projectile, with no gravity acting upon it, and no air resistance (i.e. travelling in a vaccuum) would fit that definition.
111275
Is this a question of differential equations? Just curious because that's the only time that I have ever worked with equilibrium points...if so, the equation seems to be written incorrectly. It seems confusing to have x=-3x^2-x+2 because I would just rewrite it as: 3x^2+2x+2=0. If the left side is a function it should either be y(x) or preferably, dx/dt, or some x' value. Then knowing the initial condition x(0) would make sense. With these details, I would be able to calculate the equilibrium points and classify them. You would have x=2/3 and x=-1 as your two equilibrium points, and x=2/3 would be stable and x=-1 would be unstable. Hope this helps!
981852
You can solve this problem by resolving the forces along the ramp for part a) and normal to the ramp for part b).\n\na) The net force along the ramp must be zero for the crate to move at constant velocity (ie no acceleration), so we have:\n\nFcos30 = mgsin30\nF=mgtan30\n\nb) the reaction of the ramp on the block will be normal to the ramp and equal to the sum of the imposed forces resolved normal to the ramp, so:\n\nR = Fsin30 + mgcos30
976540
a) +1\nb) 0   (in molecule state)\nc) 0\nd) -3\ne) 0\nf) +4 of Carbon\ng) +2 of Mg\nh) +1 of Chlorine (interesting!)\ni) +6 of Sulphur
912606
The simple answer is that the vector sum of the forces acting on an object must be zero if the object is to remain in "equilibrium" or in other words to have constant velocity.
1082511
I think your question have a typo.  There are three variables and I am not sure which one are you trying to solve for.
375635
Kp=Kc(RT)n\na.Kp=Kc(RT)-1/2\nb.Kp=KcRT\nc.Kp=KcRT\nd.Kp=Kc(RT)2
856026
Use the gas law\nPV=nRT
844168
This is very simple\n\nCa+H2O -> Ca(OH)2 + H2\n\nThe reactant has\n1 atom of Ca\n1 atom of H\n2 atoms of O\n\nThe product has\n1 atom of Ca, ok\n2 atoms of H\n2 atoms of O\n\nJust put a coefficient of 2 in the water (H2O)\n\nSo the chem. eqn becomes\n\nCa+2 H2O -> Ca(OH)2 + H2\n\nThat's it.\nThe equation is already balance
769637
See the link below for one explanation of "equilibrium solution" in differential equations.\n\nMy sense is that the terms you mentioned refer to different things; "lost solution" brings to mind a potential solution to a differential that may have been 'lost' during, say, division of an entire equation, or perhaps by restrictions placed on the solution set (or domain).\n\nI hope this helps pinpoint what you are after.