Professional case in the morning; I can’t remember the options

Question 6 in the professional case study in the morning – I believe such a question was indeed there. The key to solving this problem lies in the conversion of ethanol into diethyl ether through the reaction 2C2H5OH → C2H5OC2H5 + H2O (2). The total number of moles remains unchanged in this reaction. Only in the reaction for the synthesis of ethanol, C2H4 + H2O → C2H5OH (1), does the total number of moles change. As per reaction (1), for every 1 mol of C2H5OH produced, the total number of moles decreases by 1 mol. If 100 mol of reactants are used, then the total number of moles decreases by 54.5 × 6% × 92% = 3 mol. The amount of ethylene remaining is 54.5 × (1 – 6%) = 51.23 mol. Thus, the mole percentage of ethylene in the products is 51.23 ÷ (100 – 3) = 52.81%. Furthermore, if reaction (2) is considered, along with the inert gases present in the reactants other than ethylene and water, and an accounting is done for each one individually, a result can certainly be obtained as well, but I believe it would be difficult to solve this within 10 minutes.

I did the same thing during my exam. Now I have a question: the conversion rate of ethylene is 6%, with a selectivity of 92%. This question is very similar to one from the afternoon case study. Here, by selectivity, what is meant is the proportion of the reacted ethylene that remains as ethanol out of the total amount of ethylene that was converted. Looking at the equation given in the question, it seems that all of the converted ethylene ends up as ethanol, while part of the ethanol is converted into diethyl ether; this is where selectivity comes into play. The number of moles remains unchanged before and after the second reaction, so the amount that decreases is equal to the amount of ethylene that was converted, which is 54.5 x 6% = 3.27 mol. Therefore, the mole fraction of ethylene in the products is 51.23 / (100 – 3.27) = 52.96%. I remember that this value was one of the options given at that time. So this question is likely to lose points during reevaluation

This problem does require a fair amount of calculation; I worked on it in reverse, skipping the first five questions.

This post was last edited by xbl240 on 2016-9-14 09:19. There is also a question related to selection in the case study from this afternoon; I don’t quite understand what the examiner means by \"selection.\" In previous questions of this type, it involved parallel reactions such as A+B=C and A+D=F, with C being the desired product – in such cases, we wanted A to have a higher selectivity for producing C. As in the case of the selectivity mentioned in the question, all of the ethylene was actually converted into ethanol; some of the ethanol produced underwent side reactions, but everything that resulted from the conversion of ethylene was turned into ethanol. I understand that the selectivity is 100%. In reference books, selectivity is defined as the amount of reactant consumed to produce the desired product divided by the amount of reactant that is converted. So, in cases like those in the exam questions where ethanol is converted into diethyl ether, isn’t that ethanol considered a desired product? In such situations, it is the yield that is affected

The view from the 3rd floor makes more sense; the decrease in the total moles should be 54.5×6% = 3.27 mol, not 54.5×6%×92% = 3 mol. It seems I’ll lose points on this question again

For question 6 in the professional case study in the morning, I will provide the complete solution process: C2H4 + H2O = C2H5OH (1); 2C2H5OH → C2H5OC2H5 + H2O (2). By multiplying reaction (1) by 2 and adding it to reaction (2), we get 2C2H4 + H2O = C2H5OC2H5 (3). According to reaction (1), when 1 mol of C2H4 reacts, the total number of moles decreases by 1 mol. According to reaction (3), when 1 mol of C2H4 reacts, the total number of moles decreases by 1 + 0.5 – 0.5 = 1 mol. In general, for both reactions (1) and (3), the total number of moles decreases by 1 mol per 1 mol of C2H4 that reacts. Assuming 100 mol of reactants are used, 54.5 × 6% = 3.27 mol of ethylene reacts, resulting in a decrease of 3.27 mol in the total number of moles. The remaining amount of ethylene is 54.5 × (1 – 6%) = 51.23 mol. Therefore, the mole percentage of ethylene in the products is 51.23 ÷ (100 – 3.27) = 52.96%. Incidentally, if the reactions that occur are understood to be only (1) and (3), then both the conversion rate and selectivity become easy to understand, eliminating the confusion that might arise from thinking that the selectivity should be 100%! Furthermore, if reaction (2) is considered, along with the inert gases present in the reactants other than ethylene and water, and an account is taken of each one individually, a result can certainly be obtained as well; but I believe it will be difficult to solve this within 10 minutes

The independent reactions are only 1 and 2; in these reactions, everything is converted into the alcohol. Selectivity, on the other hand, refers to the formation of the desired product, rather than any remaining by-products of that desired product

Friend on the 9th floor, if you persist in trying to argue with the question-setting team in this way and fail to highlight this point, I predict that you will be at a loss regarding the two selective questions this year.

The last edit to this post was made by xbl240 on 2016-9-16 at 12:57. The question bank does not provide answers; anyway, they’re not clear, and some of the questions are indeed inaccurate. There’s that question about calculating flow rate in parallel pipelines… It’s not a big deal if there are mistakes; there have been mistakes in past exam papers – in 2013 and 2014, there was a question regarding the calculation of isentropic expansion work