Parami143_Kella 0 Report post Posted August 8, 2009 (edited) Hi! I do chemistry HL and in exams i always get the order of reaction correct but i don't get the reason correct. So could somebody please help me understand the reason for obtaining a certain order of reaction. Thank you. I have attached a past paper. I was doing the first question. Edited August 11, 2009 by Mahuta ♥ I have removed the past paper from your post, if you had a problem on the first question, then just attach a picture of the 1st question rather than the whole past paper. :) Share this post Link to post Share on other sites
sweetnsimple786 421 Report post Posted August 8, 2009 I think you could give a system of equations and solve it mathematically or write how you know it intuitively. But I think the first way is safer. So you know that Rate_{1} = (constant, k)([NO]^{x})([br_{2}]^{y} x and y are exponents and the orders that are unknownNow plug in for rate 1 and do the same for rates 2 and 3. 3.20 × 10^{-3} = (k)(2.00 × 10^{-2})^{x}(5.00 × 10^{-3})^{y} [that's rate 1]1.60 × 10^{-3} = (k)(2.00 × 10^{-2})^{x}(2.50 × 10^{-3})^{y} [rate 2]Now to find y, the rate order for Br_{2}, divide equation one by equation 2. The k and the [NO] cancels out, giving you.00320/.00160 = (.005/.0025)^{y}2 = 2^{y} so y = 1You'd do the same to find x, except now you use rates 1 and three [because to find x [NO] should change and it's easier when [br] stays the same, even though you can find the second order after you have the first one]3.20 × 10^{-3} = (k)(2.00 × 10^{-2})^{x}(5.00 × 10^{-3})^{y}1.30 × 10^{-2} = (k)(4.00 × 10^{-2})^{x}(5.00 × 10^{-3})^{y}Now divide equation 3 by equation 1. The k and the [br_{2}] will cancel out leaving.013/.0032 = (.004/.002)^{x}4.0625 = (2)^{x} and you can take the log of both sides to find x, but you know the order is 2 [you round a bit] so x = 2 Share this post Link to post Share on other sites
Sandwich 2,417 Report post Posted August 8, 2009 For a 2 mark question in an exam, I believe what they're after is something along the lines of "doubling the concentration of NO increased the rate by a factor of 4, and doubling the concentration of Br2 increased the rate by a factor of 2, therefore the NO is evidently involved twice before the rate determining step and is raised to a power of 2, and the Br2 involved only once, so raised to the power of 1."Or something like that. They're more or less just after proof you did it by recognising the relationship between change in rate as it correlates with change in concentration, and some proof you know how that is going to change the powers. At least that's what I recall from back when I did past papers!Out of curiosity, how can you work out the order of reaction without knowing the reason? Share this post Link to post Share on other sites
Parami143_Kella 0 Report post Posted August 8, 2009 (edited) Well our teacher told us to look at the rate of reaction then look at the concentration and see how the rate increases or decreases with the concentration. That's why i got confused. But thank you for your help. Thank both of you. Edited August 8, 2009 by Parami143_Kella Share this post Link to post Share on other sites
Sandwich 2,417 Report post Posted August 8, 2009 Well our teacher told us to look at the rate of reaction then look at the concentration and see how the rate increases or decreases with the concentration. That's why i got confused. But thank you for your help. Thank both of you. ^^ Really, how you worked it out is all you have to put as your explanation! Share this post Link to post Share on other sites