Texas Tech University
The Integrated Rate Laws - Example 4


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So there's a question about the second order reactions. So our X Y reacting is going to some products in a second order. Fashion The question they're giving us, uh, the K constant 7.2 times 10 to the part of negative tree loyalty to the port of native one seconds to the power of negative one. If you can recall, uh, the previous lectures, this is three units for a second order, the composition reactions k constant. And our initial concentration is zero point 15 and our final concentration is going to be 0.6 They're asking this time required to bring the initial concentration to do final concentration. So what we need to do is to write Thesiger and order, uh, integrated rate law one over X. Why T equals two Casey plus one over X Y zero. Now, if you want to put these numbers in, this is going to be one over zero point 06 equals two some 0.2 times 10 to the part of negative tree T plus one over 0.15 year, and we're going to have 1/0 0.6 minus 1/0 0.150 divided by same 0.2 times 10 to the power of negative three equals City and T will be equal to well for 34.50 seconds.

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