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I tried finding the slope between two points but I keep getting wrong values, how does one find the initial rate by using such graphs?

So I started by trying to find a slope of two points, $(0,0.4)$ and $(10,0.2)$ after calculations I got $-0.02$, which is the opposite of the right answer in sign. I tried to check myself by trying to find the slope for other points like: $(10,0.2)$ and $(20,0.1)$ which gives me $-0.01$, I'm confused as whether my initial idea is right or not.

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  • $\begingroup$ This standard exam question is a first order decay with half life (from the graph) of about 13 s. So the rate constant, k, is $(ln2)/t_{1/2}$, which is about $0.053 s^{-1}$. Hence the initial slope is -k times 0.4 M, giving $0.021 M s^{-1}$ as the magnitude of the initial slope. Rounded to one significant figure, this is $0.02 M s^{-1}$, as per answer A. $\endgroup$ – Ed V Feb 24 at 17:29
  • $\begingroup$ @EdV Well could you elucidate on the fact how you found out the half-life to be of 13s from the graph, I mean that's exactly I've been searching for! Is this a previously done question? I mean yeah it's an exponential curve but how are we being so sure of it? $\endgroup$ – Sir Arthur7 Feb 24 at 17:55
  • $\begingroup$ @EdV Well I was actually asking how are you so sure about the coordinates? The graph lines don't intersect the graph exactly at any point other than at $(0,0.4)$, there's always a chance that $t_{{1/2}}$ =14s instead of 13 right? And that's exactly why one can't be so sure that the $t_{{1/2}}$ is constant or not. Being an exam question from your previous knowledge, you're probably right, beginners aren't made to fuss, they stick to order 0,1 or at max 2. But we should try to be general whenever possible, for e.g. here nothing's been mentioned! $\endgroup$ – Sir Arthur7 Feb 24 at 18:49
  • $\begingroup$ @EdV No point in arguing, I just personally feel that those dashes don't count as a scale of measure, but that doesn't change the answer or the method itself. Since you've been teaching such standard exam questions, you're surely better than me here. Well you could actually put this up as an answer, people could upvote it! $\endgroup$ – Sir Arthur7 Feb 24 at 19:05
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It seems like you have the right idea to me,

For the "Initial Rate," you're taking a curve and doing a straight approximation of the curve over a small portion, as you appear to be doing.

Your logic also appears correct to me, a decrease in concentration should correspond to a negative rate.

My guess is that the question, (a poorly worded one?) is actually referring to the rate of product creation, and asking you to assume that there is a 1-1 relationship between the decrease in reactant, and the appearance of product. Thus if your reactant is disappearing at $0.02~M/s$ (as indicated by $rate= -0.02 ~M/s$) , then your product should be appearing at $0.02~M/s$, making A the correct answer.

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  • $\begingroup$ yeah, but since it's a decelerating curve it wont have the same slope all the time right? how am I to find the right one if the slope is always different... $\endgroup$ – Jx1 Jun 9 '15 at 3:06
  • $\begingroup$ I agree, the question in the textbook seems very vague $\endgroup$ – user15489 Jun 9 '15 at 4:16
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    $\begingroup$ @Jx1 That's correct! That's where the approximation and initial keywords come in. You're approximating the slope at a given point, meaning that you're using a line that has about the same slope as the curve over a specific region. Initial indicates that this region as at the very beginning. (The initiation of the reaction.) So you're drawing a straight line that's the same as the line at the very beginning of the reaction. $\endgroup$ – Dan Jun 9 '15 at 14:17
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This is a standard exam question pertaining to the kinetics material that is typically presented in a first year general chemistry course. Usually the coverage is restricted to discussion of zero, first and second order kinetics and this is what gets tested on exams. From the photo of the exam question, the decay is apparently first order, i.e., $$\mathrm C(t) = C(0)e^{-kt}$$ where $\mathrm C(t)$ is the molar concentration of whatever is decaying away, $\mathrm C(0)$ is the initial molar concentration (i.e., at $\mathrm t = 0$) of whatever is decaying away, and $k$ is the decay rate constant in units of $\mathrm s^{-1}$. In the present case, $\mathrm C(0) = 0.4 M.$

But, is the decay really exponential? To determine this, without curve fitting data pairs or linearizing the plot, it suffices to see if the half-life, $\mathrm t_{1/2}$, is constant. From the photo of the graph, the initial concentration declines to $\mathrm 0.2 M$ at about $\mathrm t = 13 s$. Note that there are $\mathrm 10$ dash marks every $\mathrm 10 s$ along the time axis, so this is an aid, albeit imperfect, in estimating times along the time axis. The concentration declines to $\mathrm 0.1 M$ at about the $\mathrm 26 s$ mark and further declines to about $\mathrm 0.05 M$ at about the $\mathrm 39 s$ mark.

So, $\mathrm t_{1/2}$ is constant and approximately equal to $\mathrm 13 s$. Therefore, the decay process is a first order exponential, as shown in the equation at the top. Note that the only other alternatives, in the restricted context of typical course material coverage and exam coverage of said material, are zero and second order decays. But these do not have constant half-lives: for a zero order decay, every subsequent half-life is halved and for a second order decay, every subsequent half-life is doubled. The graph clearly eliminates these two possibilities. It is also obvious that the decay cannot be first order since then the concentration versus time plot would be linear. If the decay was second order, the concentration could not be down to $\mathrm 0.1M$ in less than $\mathrm 30s$, even if the first $\mathrm t_{1/2}$ was clearly underestimated as $\mathrm 10s$.

The relationship between $\mathrm k$ and $\mathrm t_{1/2}$ is $\mathrm k = (ln2)/t_{1/2}$. This is easily verified by substitution into the exponential equation, resulting in $$\mathrm C(t = t_{1/2}) = C(0)/2$$ Hence, with $\mathrm t_{1/2}$ estimated as $\mathrm 13s$, the rate constant estimate is $\mathrm k = 0.053 s^{-1}$.

To find the initial rate of decay, simply differentiate the exponential equation and evaluate it at $\mathrm t = 0$. The derivative is $$\mathrm dC(t)/dt = -kC(0)e^{-kt}$$ so at $\mathrm t = 0$, the slope is simply $\mathrm -kC(0)$ and the magnitude of the slope is $\mathrm kC(0)$. With $\mathrm k = 0.053 s^{-1}$ and $\mathrm C(0) = 0.4 M$, this yields $\mathrm 0.021 M s^{-1}$ as the initial rate magnitude. Rounded to one significant figure, which is all this exam problem deserves, yields $\mathrm = 0.02 M s^{-1}$. This is exam answer A.

One last thing: a small error in estimating the half-life will not matter because the initial decay rate is inversely proportional to it, so it would have to be high by a factor of two in order to get answer B on the exam.

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  • $\begingroup$ No calculus? No problem! Since $\mathrm e^{-x} \approx 1-x$ for $\mathrm x << 1$, assume $\mathrm 0 < t$ and $\mathrm kt << 1$. Then $\mathrm C(t) \approx C(0)[1-kt] = C(0) - C(0)kt$. Subtract $\mathrm C(0)$ from both sides and divide both sides by $\mathrm t$. Then the approximate initial slope is $\mathrm [C(t) - C(0)]/t$, which equals $\mathrm -C(0)k$, as in the answer. N.B. Taking the limit, as $\mathrm t$ goes to zero, gives the true derivative. $\endgroup$ – Ed V Feb 25 at 23:42
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The idea is that you want to find the instantaneous rate of change (rise over run)at the initial value (time = 0). but since this gives .4/0 (can't be done) just use a close number to the initial value to get a good approximation of the slope at the value. So for initial so (.4-.3)/(0-5) =.02 you use the slope because the units are Moles(rise)/Seconds(run). since it doesn't specify the rate of appearance or disappearance then the magnitude is what matters not sign.

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    $\begingroup$ Welcome to Chem.SE! Thanks for posting an answer here. Please check out the tour and help center for more information on the site and our policies. In particular, please take a look at the page on formatting posts; yours is a bit hard to read at the moment. Also, once you reach 20 rep, please drop by chat and say hello! $\endgroup$ – hBy2Py May 3 '17 at 4:13
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Firstly, you're right that the slopes at all points on this curve will be negative as it's a downward sloping curve. The rate of reaction w.r.t. reactants is the negative of the rate of consumption of reactants provided there's 1:1 stoichiometric ratio between the reactants and products. But in this question, nowhere is it mentioned that this curve depicts the concentration vs time curve for reactants. Even if we consider it to be so, there's no mention of a function that describes this graph, from which we could have found out the order and rate constant of the reaction.

Thus we have to proceed in a way @Dan pointed out, you need to use the slope a line that has approximately the same slope as the curve at the initial state of the reaction. But for that we need 2 exact coordinates of very closely spaced points on this graph, just like you tried to do, but as you see, the graph doesn't pass through $(10,0.2)$, actually there's no other point than $(0,0.4)$ whose coordinates we know for sure. But if you look closely, you'd find $(5,0.3)$ is a better match (as far as my eyes tell me). So using that, one can find the slope of line between them to be $\frac{0.3-0.4}{5-0}=-0.02$

Thus the rate of reaction w.r.t. reactants could be-

Rate $=\frac{\mathrm{-d}[\text{reactants}]}{\mathrm{d}t}=-(-0.02)=0.02 \pu{mol L^-1 s^-1}$

Even though my stand is that this is a poor question that lacks a lot of necessary detail. So it's better not to waste time on this one, but yeah there's always a chance of learning something out of everything.

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  • $\begingroup$ @EdV You're welcome :) Yeah actually I found that "Don't write EDIT" part from that post itself. I prefer that post over all other formatting posts, because it's mostly community agreed. But you could have posted this comment on your post too, that would have been better right? $\endgroup$ – Sir Arthur7 Feb 24 at 17:58
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    $\begingroup$ @EdV You're right, officially we are supposed to avoid such things, but it's mostly confined to Q&A, where we should avoid "Thanks for any help" or "Hope that this helps", but informally we're quite lax on doing that in comments, until it leads to an extended discussion (and eventually moved to chat). Chem SE is a small community and we mostly know each other, so we're kinda family at times, we thank others if they find out a typo in our posts, or give some advice on formatting. $\endgroup$ – Sir Arthur7 Feb 24 at 18:59

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