For the cases from the afternoon session of 2016’s Note: Chemical treatment course, seeking conversion rates; selective approach
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I can’t remember the title. Reaction (1) is A+B=P, and reaction (2) is A+P=S. It is given that before the reaction Ca0=3.0 and Cb0=5.0; what is the value of Ca after the reaction? Cb=? Cp=? Find, based on B, B’s selectivity and P’s yield... Do everyone still remember the details? During the calculation, I remember there was a concentration related to Ca around 2.3; due to time constraints, I wasn’t able to finish the calculations. I completed only the first step, wrote a formula for the final yield, and submitted my answer by guessing
I remember it was Ca=0.4 and Cb=2.7. My solution process was as follows: Reaction (1) plus Reaction (2) yields 2A+B=S (3). Assuming a volume of 1 m³, the amount of A that reacted was 3–0.4 = 2.6 Kmol, while the amount of B that reacted was 5–2.7 = 2.3 Kmol. Let x be the amount of B that reacted according to Reaction (1), and y be the amount of B that reacted according to Reaction (3). Then x + y = 2.3, and X + 2y = 2.6. Solving these equations gives x = 2 and y = 0.3. Thus, the selectivity is 2/2.3 = 87%, and the yield is 2/5 = 40%
Thank you; it was a very simple question. I really didn’t have time for the second step, so I submitted my answer. I got 87% correct; for the remaining two options, I chose the wrong one, haha
Oh, I didn’t combine the two reactions; I calculated them separately, and that also gave 0.87. If I use your method, I can save half the time, and I’ll have time to do the second step as well… Another 2 points lost