Chem equilibrium problem. Need help fast!?
(a) Kc = [CO] [H2O] / [CO2] [H2]; Kc = 0.050 * 0.040 / 0.086 * 0.045 = 0.517. (b) If [CO2] is quickly changed to 0.25 M, then the equilibrium will respond by shifting to the right. CO2 & H2 will both lose x M in concentration, and CO and H2O will gain x M in concentration. Placing the equilibrium concentrations into the Kc expression, we obtain: 0.517 = (0.050+x) (0.040+x) / (0.25-x) (0.045-x). Rearranging into standard quadratic form, we obtain: 0.483x² + 0.242x - 0.00381 = 0 Using the quadratic formula, the only positive x for this equation is 0.015 M. The equilibrium concentrations are then: [CO2] = 0.25 - 0.015 = 0.235 M [H2] = 0.045 - 0.015 = 0.030 M [CO] = 0.050 + 0.015 = 0.065 M [H2O] = 0.040 + 0.015 = 0.055 M. If these new equilibrium concentrations are used to recalculate Kc, the same value is obtained and so these answers check.
Chem equilibrium problem. Need help fast!?
2 SO2 + O2 => 2 SO3 Kc= [SO3]^2 /( [O2][SO2]^2 ) S(s) + O2(g) SO2(g) K'c = 4.2 1052 (*) 2 S(s) + 3 O2(g) 2 SO3(g) K"c = 9.8 10128 (**) from (*) you get [SO2]/ ([O2]*[S]) = 4.2 1052 = K'c from (**) you get [SO3]^2 / ([O2]^3 *[S]^2 ) = 9.8 10128 = K''c from (*) [SO2] = K'c*[O2]*[S] then [SO2]^2 = K'c^2*[O2]^2[S]^2 then [SO2]^2[O2] = K'c^2[O2]^3[S]^2 from (**) [SO3]^2 = K''c*[O2]^3*[S]^2] the problem is solved (do the ratio) [SO3]^2/( [SO2]^2*[O2]) = Kc Kc = K''c/(K'c^2)
Consider the following reaction at equilibrium.. help fast? thanks?
Kc= [H2S]²/([H2]²*[S2]) a) c= n/V= m/(M*V) ; c: concentration; M: molecular mass V: volume; n: number of mol. M(H2S)=34,06 g/m0l, M(H2)= 2g/mol, M(S2)=64.12 g/mol, V= 10.0 L -> [H2]=0.10 M, [S2]=0.016 M, [H2S]=0.199 M -> Kc= 248 1/M b) [S2]= (1/Kc)*([H2S]/[H2])² ; V=5.0L [H2]= 0.06 M(=n/V); [H2S]=0.5 M -> [S2]= 0.280 M Check: With the all the concentrations of b) Kc becomes 248 1/M
Does a high equilibrium constant make a reaction move faster?
No, as the equilibrium constant only tells what the concentrations of the reactants and products are at equilibrium. The constant that determines the speed of a reaction is the rate constant.
Chemistry Equilibrium problem, Help Fast pls. Thanks to all those who do =D?
initial concentration SO3 = 2/8.00 L=0.25 M concentration at equilibrium of O2 = 0.58 / 8.00=0.0725 M 2 SO3 <==> 2 SO2 + O2 initila concentration 0.25 change -2x .. . .. . .. . . +2x .. . .+x x = 0.0725 2x = 0.145 [SO3]= 0.25 - 0.145 =0.105 M [SO2]= 0.145 M [O2] = 0.0725 Kc = [SO2]^2 [O2] / [SO3]^2 = (0.145)^2 ( 0.0725 / (0.105)^2 = 0.138 0.00609 - x = 0.00301 x = [COCl2] = 0.00308 M [CO]= 0.0102 - 0.00308=0.00712 M Kc = 0.00308 / 0.00712 x 0.00301 = 143.7
AMAZING EQUILIBRIUM PROBLEM! PLS HELP.?
It would help if you would say what you think you need to solve a problem that you don't already have. Often by doing so you'll see that you do have everything you need and really do know how to solve it. This is a closed vessel. None of the players are going anywhere else; if HI is disappearing, then it must be showing up as H2 and I2. Otherwise, you're about to experience the event horizon first-hand. (1.0-0.8 mol HI) / (1 L) . . . [which is the HI that disappeared] x (1 mol H2) / (2 mol HI) . . . [stoichiometry] Pro forma I2; [H2] = [I2] = 0.1M, [HI] = 0.8M Kc = [H2][I2] / [HI] . . . [which is arithmetic] Kp = pH2 pI2 / pHI You might memorize the relationship between Kc and Kp, but I prefer to retrieve it from the ideal gas law, pV = nRT by recalling that n/V is the gas concentration: Kp = { n/V RT }_h2 { n/V RT }_i2 / { n/V RT }_KI = Kc (RT)
CHEMISTRY partial pressures equilibrium problem, plz help?
First, calculate Kp for the original equilibrium: Kp = (PHI)^2 / PI2 PH2 = 0.1^2 / (0.2)(0.2) = 0.25 So, as soon as the additional HI is added, the initial pressures of each gas are: H2 = 0.2 I2 = 0.2 HI = 0.32 Some of that added HI will decompose to form more H2 and more I2. So, let x = atm of H2 and I2 formed. Then, because of the balanced equation, 2x atm of HI will disappear. So, the final pressures of each gas will be: H2 = 0.2 + x I2 = 0.2 + x HI = 0.32 - 2x Now, plug those back into the expression for Kp to give: 0.25 = (0.32-2x)^2 / (0.2 +x)^2 Simplify this by taking the square root of each side first: 0.5 = (0.32 - 2x) / (0.2 +x) Now, you've got a fairly basic algebra problem to solve. Remember that when you get to x, you can calculate the equilibrium pressure of HI as 0.32 - 2x. Good luck and hope this helps...