Методическая разработка по английскому языку для студентов факультета химии и химической технологии " Simple Chemistry" Алматы 2017
Example 3: The oxidation of ethanol by acidified potassium dichromate(VI)
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| Example 3: The oxidation of ethanol by acidified potassium dichromate(VI)
This technique can be used just as well in examples involving organic chemicals. Potassium dichromate(VI) solution acidified with dilute sulphuric acid is used to oxidise ethanol, CH3CH2OH, to ethanoic acid, CH3COOH. The oxidising agent is the dichromate(VI) ion, Cr2O72-. This is reduced to chromium(III) ions, Cr3+. We'll do the ethanol to ethanoic acid half-equation first. Using the same stages as before, start by writing down what you know: Balance the oxygens by adding a water molecule to the left-hand side: Add hydrogen ions to the right-hand side to balance the hydrogens: And finally balance the charges by adding 4 electrons to the right-hand side to give an overall zero charge on each side: The dichromate(VI) half-equation contains a trap which lots of people fall into! Start by writing down what you know: What people often forget to do at this stage is to balance the chromiums. If you don't do that, you are doomed to getting the wrong answer at the end of the process! When you come to balance the charges you will have to write in the wrong number of electrons - which means that your multiplying factors will be wrong when you come to add the half-equations . . . A complete waste of time! Now balance the oxygens by adding water molecules . . . . . . and the hydrogens by adding hydrogen ions: Now all that needs balancing is the charges. Add 6 electrons to the left-hand side to give a net 6+ on each side. Combining the half-reactions to make the ionic equation for the reaction What we have so far is: What are the multiplying factors for the equations this time? The simplest way of working this out is to find the smallest number of electrons which both 4 and 6 will divide into - in this case, 12. That means that you can multiply one equation by 3 and the other by 2. The multiplication and addition looks like this: Now you will find that there are water molecules and hydrogen ions occurring on both sides of the ionic equation. You can simplify this to give the final equation: Literature Н.А. Степанова. Practical course of English. About the foundation of chemistry. Спб.: 2011, Политехника. – 120 с. Mark Ibbotson. Professional English in Use. Engineering. Cambridge University Press, 2011, p. 144. Michael McCarthy, Felicity O’Neil. Academic Vocabulary in Use. Cambridge University Press, 2012, p. 176. A Dictionary of Chemistry. Fifth Edition. Oxford University Press. 2004, p. 602. General Chemistry.// wikipedia. com Norman Coe, Mark Harrison, Ken Patterson. Oxford Practice Grammar. Oxford University Press, 2006, p. 269. http://www.perfect-english-grammar.com http://www.talkenglish.com/grammar www.wikipedea.com Kenna Bourke. Grammar. Pre-Intermediate. Oxford University Press, 2010. p.87 Michael Swan. Practical English Usage. Third Edition. Oxford University Press, 653. жүктеу/скачать 422,31 Kb. Достарыңызбен бөлісу:
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