Question 5 in the morning session of the 2013 case
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5. The final mixture resulting from the chlorination of benzene contains 39% (by weight) of chlorobenzene (C6H5Cl), 1% of dichlorobenzene (C6H4Cl2), and 60% of benzene (C6H6). The known reaction equations are:C6H6(l) + Cl2(g) → C6H5Cl(l) + HCl(g) (1)
The enthalpy change for this reaction is -193.13 kJ/kmol.
C6H6(l) + 2Cl2(g) → C6H4Cl2(l) + 2HCl(g) (2)
The enthalpy change for this reaction is -286.64 kJ/kmol.
Molecular weights: C6H6 = 79, C6H5Cl = 112.5, C6H4Cl2 = 147. What is the enthalpy change in kJ per ton of benzene during this reaction? (A) -7868 (B) -3934 (C) 2221 (D) 5650 Is there a problem with the numbers in this question as they were copied from somewhere else? I just can’t figure out the answer no matter what! Seeking answers from experts. I calculated it to be -39322; the difference from that is huge!
5000/(60+39*79/112.5+79/147)*(39/112.5*193.13+1/147*286.64)=3918 kj; calculation error.
I did the calculations; I made a mistake with the decimal points. It should be =5000/79/(39/112.5+1/147+60/79)(39/112.5*193.13+1/147*286.64)=3918
Using 1 ton of raw material as the basis for calculation: it contains chlorobenzene (390 kg), dichlorobenzene (10 kg), and benzene that does not participate in the reaction (600 kg). Thus, the heat energy Q = (390/112.5)*(-193.13) + (10/147)*(-286.64) = -689.02 KJ. The amount of benzene that participates in the reaction is 600 + (390/112.5)*78 + (10/147)*78 = 875.71 kg, which is equivalent to 0.8757 tons. In other words, 0.8757 tons of benzene can release 689.02 KJ of heat. So, how much heat can 5 tons of benzene release? Thus, Q = -689.02*5/0.8757 = -3934 KJ; the answer is (B) – it’s an easy question! Done in 4 minutes!
Technicians are really amazing; no wonder the answer matched the correct one. It turns out that for the question about the molecular weight of benzene, the answer given was 79. .
-3950: First, the mole fractions of chlorobenzene, dichlorobenzene, and benzene are calculated based on their mass fractions. The mole fraction of chlorobenzene corresponds to the conversion rate of benzene in the first reaction, while the mole fraction of dichlorobenzene represents the conversion rate of benzene in the second reaction. From these values, the amount of benzene consumed in each reaction can be determined, and subsequently, the heat of reaction for each reaction can be calculated.
Technician Zhang has a clear approach to solving problems and uses simple methods, which is worth learning from*. It’s just that the pressure is a bit high.
It’s okay; I’ll help you answer your questions while working! I’ll do my best to broaden your thinking!
The problem here stems from the incorrect molecular weight of benzene! 78 was written as 79; the answer is correct only if that’s the case, damn it! :@
The molecular weight of benzene given in the question is incorrect; it should be 78! Otherwise, answer B cannot be calculated! What an idiot who came up with those questions