CO(g) + 3H2(g) <==> CH4(g) + H2O(g)... AT 1127 Deg. C. analysis at equilibrium s

Question

CO(g) + 3H2(g) <==> CH4(g) + H2O(g)... AT 1127 Deg. C. analysis at equilibrium shows the presence of 0.0321 mol CO, 0.0421 mol H2, 0.00136 mol CH4, and 0.000432 mol H2O in a 4.00 liter container... A) CALCULATE Kc FOR THIS REACTION AT 1127 DEG. C. ... ... B) CALCULATE Kp FOR THIS REACTION AT 1127 DEG. C... MY ANSWERS AS FOLLOWS: (0.00136mol/4.0L) x (.000432/4.0L) / (0.0321/4.0L) x (0.0421/4.0L)^3 ... ---> ... [ (0.0034)x(0.000108) ] / [ (0.008025) x (1.2x10^-6)^3 ] ... ---> ... (3.672x10^-8) / (1.38672x10^-20) = 2.6x10^12 = Kc... IS THIS CORRECT...

Explanation / Answer

here let me give you my solution as per the question we calculate the total pressure of the system using ideal gas equation ie P V = n R T where all terms carry their usual meanings so now V=4 l n is sum of all moles = 0.0759 R = 0.0821 T = 1400 k calculating we get P = 2.81 atm so partial pressure of each sample is Pco = mole fraction * total pressure ie .0421/.0759 * 2.81 = 1.188 similarly PH2 = 1.55 PCH4 = .05 PH20=.015 atm so Kp= PH2O*PCH4/PH2^3 * PCO = 1.706 * 10^-4 and relation between kp and kc is kp = kc*(RT)^delta n ie difference between no of moles of products- reactants here delta n = -2 ie kc = kp * (RT)^2 = 2.25

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