. The equilibrium constant for the reaction CO (g) + H2O (g) ---> CO2 (g) + H2 (g)?
Kc = [CO2][H2] / [CO][H2O] From ICE table, you can obtain this equilibrium expressions: [CO] = 3 - x [H2] = x [H2O] = 1 - x [CO2] = x Solve for x: 0.63 = (x)(x) / (3-x)(1-x) x = 0.6818 moles = [H2] =====> answer in letter a ______________________________________... @ equilibrium: [CO] = 3 - 0.6818 = 2.318 moles [H2] = 0.6818 mole [H2O] = 1- 0.6818 = 0.3182 mole [CO2] = 0.6818 mole total moles of gas @ equilibrium = 4 moles To calculate for the partial pressure of the gas, multiply the total pressure with the mole fraction of the gas: par. P of CO = (2.318 mol / 4 mol)(2 atm) = 1.159 atm ===> answer par. P of H2 = (0.6818 mol / 4 mol)(2 atm) = 0.3409 atm ===> answer You continue the rest. ______________________________________...
Estimate the value of the equilibrium constant at 690 K for the following reactions. ΔH∘f and S∘ for BrCl(g) is 14.6 kJ/mol and 240.0 Jmol⋅K?
ΔGf = ΔH∘f -TΔS∘f = 14.6 * 10 ^3 kJ/mol - 690 K* 240.0 Jmol⋅K^-1 =-151000 j mol ^-1 k = exp( - delta G / RT) = exp ( -151000 / 8.314 * 963) = 6.44004 *10 ^-09
The activation energy for the reaction 2HI --> H2 + I2 is 184 kj/mol.?
Arrhenius equation: 184 kilojoules = 184,000 kilojoues ; T1 = 500 + 273 = 773 K; T2 = 520 + 273 = 793 K ln k2/k1 = Ea/R(1/T1 - 1/T2) ln k2/k1 = 184,00/8.314( 1/773 - 1/793) = 22,131 ( 0.001294 - 0.001261) = 22,131( 0.000033) ln k2/k1 = 0.730 e^0.730 = 2.07 k2/k1 = 2
HELP WITH A CHEM PARTIAL PRESSURE AT EQUILIBRIUM QUESTION!?
In addition to the H2S already present in the flask, solid NH4HS is added until there is excess unreacted solid remaining. The initial pressure of H2 in the flask is 0.0574atm. a) What are the partial pressures of NH3 and H2S at equilibrium, that is, what are the values of PNH3 and PH2S, respectively? I've done everything literally and no answer appears to be correct. I know that I need to find x for NH3 and x+0.0574 for H2S. Also, b) What is the mole fraction, χ, of H2S in the gas mixture at equilibrium? c) What is the minimum mass of NH4HS that must be added to the 5.00-L flask when charged with the 0.400g of pure H2S(g), at 25 ∘C to achieve equilibrium? Thank you!
How would you calculate the value of the equilibrium constant, Keq , for the following reaction at 298 Kelvin?
Calculate the value of the equilibrium constant, Keq , for the following reaction at 298 Kelvin. (Use the reaction free energy given below.) 2SO3 = 2SO2 + O2 in the gaseous state ΔGo = 140.0 kJ/mol any help would be greatly appreciated~!
Chemistry help please!! (equilibrium)?
a) Keq= [ products ] /[ reactants] Keq= [CO2] * [H2] / [H2O] *[CO] so you plug in the value however your value are in mol and not expressed as concentration so C=n/V for all the value for instance [CO]= 0.3/1 = 0.3 mol.L-1 Keq= (0.5 * 0.5) / (0.3* 0.3) =2.78 their value is 10 and the one we found is 2.78 which doesnt math so the system is not at equilibrium b) THE answer in a is actually Qeq ( another name for Keq) so here Keq>Qeq when you have Keq bigger than Qeq, the reaction will shift to the right ( just a tip to remember look at the arrow it point to the right , make sure you write K before Q) Therefore more products will formed as it is a right shift c) So if you do an ICE table it will be easier ________________CO(g) + H2O(g) ⇔ CO2(g) + H2(g) Initial concentration 0.3_____ 0.3 ______ 0.5___ 0.5 Concentra change - x_____ - x _______ + x ___+ x Equation (I+C line) 0.3-x_____0.3-x _____ 0.5+x __0.5+x so you put that in your equation Keq= [CO2] * [H2] / [H2O] *[CO] 10=((0.5+x) *(0.5+x)) / (0.3-x) * (0.3-x) x=0.107798 so you plug that in your value [CO]=[ H2O]=0.3-x =0.1922mol.L-1 [ CO2]= [H2]= 0.5+x = 0.6078 mol.L-1 I finally finished !! if you are not familiar with ICE table you can google it, they will show you how to do it cauz its really hard to line the value with yahoo answers