Guest Posted January 15, 2016 Report Share Posted January 15, 2016 Please help me understand this question! Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 15, 2016 Report Share Posted January 15, 2016 (edited) A mechanism is a schematic understanding of a complex, multi-step reaction by chunking it into simpler, often involving two reactants, reactions. The mechanism is presented in order of precedence. In this case, the slow step occurs before the two fast steps.Recall what steps in a mechanism contributes to the rate law. Write the rate law.For explanatory purposes, suppose the rate is proportional toA*B^2*C^3*D^4 If you double the concentration for each species given in the choices, {or example for choice a) you only double A, for choice b) you double A and double C, and so on} which combination causes the greatest change in rate?Just saying, you shouldn't eliminate any choices before you understood the question!EDIT: some text-rendering issues. I meant concentration for any referals to A, B, C, D Edited January 15, 2016 by kw0573 Reply Link to post Share on other sites More sharing options...
Guest Posted January 15, 2016 Report Share Posted January 15, 2016 (edited) Ok, but in the first equation, there are 2 mols of A and C. So shouldn't the answer be B? Since increasing the concentration will affect the rate?(the answer is A, by the way. I relaized that I hadn;t mentioned that.) Edited January 15, 2016 by frank!e Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 15, 2016 Report Share Posted January 15, 2016 (edited) For a non-equilibrium slow reaction, the rate is just the reactant side with the coefficients as the exponentsr = k * A^2It's a bad question because it should be "substance(s)" in the question. And a, b, and e should be equally correct, because C and D do not contribute to the rate. However if you pick only b, that is not right because you hadn't gotten the right rate law. Edited January 15, 2016 by kw0573 Reply Link to post Share on other sites More sharing options...
Vioh Posted January 15, 2016 Report Share Posted January 15, 2016 (edited) It's a bad question because it should be "substance(s)" in the question. And a, b, and e should be equally correct, because C and D do not contribute to the rate. However if you pick only b, that is not right because you hadn't gotten the right rate law. Not really, because I think both C and D can contribute to the rate. In this case, they would slow down the reaction because C is the product of the first step of the reaction, and D is the product of the second step of the reaction. And by the Le Chartelier's principle, increasing the concentration of the products would shift the reactions to the left, thus slowing down the overall three-step reaction above. So I think only (a) should be the correct answer. Isn't that right? Edited January 15, 2016 by Vioh Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 16, 2016 Report Share Posted January 16, 2016 It's a bad question because it should be "substance(s)" in the question. And a, b, and e should be equally correct, because C and D do not contribute to the rate. However if you pick only b, that is not right because you hadn't gotten the right rate law. Not really, because I think both C and D can contribute to the rate. In this case, they would slow down the reaction because C is the product of the first step of the reaction, and D is the product of the second step of the reaction. And by the Le Chartelier's principle, increasing the concentration of the products would shift the reactions to the left, thus slowing down the overall three-step reaction above. So I think only (a) should be the correct answer. Isn't that right? The slow step is not an equilibrium but it goes to completion. Le Chantelier's principle only works for shifting equilibrium position. Regardless, the question can only say the slow step goes to completion, and not in equilibrium, WITH THE ASSUMPTION that the reverse reaction has no effect or is essentially not observable. We cannot break this assumption when looking at choices B and E. 1 Reply Link to post Share on other sites More sharing options...
Vioh Posted January 16, 2016 Report Share Posted January 16, 2016 The slow step is not an equilibrium but it goes to completion. Le Chantelier's principle only works for shifting equilibrium position.Regardless, the question can only say the slow step goes to completion, and not in equilibrium, WITH THE ASSUMPTION that the reverse reaction has no effect or is essentially not observable. We cannot break this assumption when looking at choices B and E. yes, you're absolutely correct. I completely forgot that the equilibrium sign is not there in the equations. ugh... such a bad question! Reply Link to post Share on other sites More sharing options...
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