47. At low to moderate pressures, the equilibrium state of the water-gas shift reaction CÓ + H;O = CO, + H is approximat

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47 At Low To Moderate Pressures The Equilibrium State Of The Water Gas Shift Reaction Co H O Co H Is Approximat 1 (41.77 KiB) Viewed 39 times

47. At low to moderate pressures, the equilibrium state of the water-gas shift reaction CÓ + H;O = CO, + H is approximately described by the relation = K.(T) = 0.0247 exp[4020/T(K)] YCO, YH₂ = YOO YH₂O where T is the reactor temperature, K, is the reaction equilibrium constant, and y, is the mole fraction of species i in the reactor contents at equilibrium. The feed to a batch shift reactor contains 20.0 mole% CO, 10.0 % CO₂, 40.0% water, and the 1123 K. balance an inert gas. The reactor is maintained at T (a) Assume a basis of 1 mol feed and draw and label a flowchart. Carry out a degree-of-freedom analysis of the reactor based on extents of reaction and use it to prove that you have enough information to calculate the composition of the reaction mixture at equilibrium. Do no calcula- tions. (b) Calculate the total moles of gas in the reactor at equilibrium (if it takes you more than 5 seconds you're missing the point) and then the equilibrium mole fraction of hydrogen in the product. (Suggestion: Begin by writing expressions for the moles of each species in the product gas in terms of the extent of reaction, and then write expressions for the species mole fractions.) (c) Suppose a gas sample is drawn from the reactor and analyzed shortly after startup and the mole fraction of hydrogen is significantly different from the calculated value. Assuming that no cal- culation mistakes or measurement errors have been made, what is a likely explanation for the discrepancy between the calculated and measured hydrogen yields?