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Excess magnesium was added to a beaker of aqueous hydrochloric acid on a balance. A graph of the mass of the beaker and contents was plotted against time (line 1).
What change in the experiment could give line 2?

  1. The same mass of magnesium but in smaller pieces
  2. The same volume of a more concentrated solution of hydrochloric acid
  3. A lower temperature

Original question:
mass vs time

The answer is just II. However, I thought that both I and II would answer the question. Why is it just II?

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    $\begingroup$ $I$ is definitely not the case. They would have converged by the time they flattened out if the same amount of magnesium was used. See the addition to Hippalectryon's answer below and the note I made there. The only way the final mass would change (out of the three given) would be if more $\ce{H2}$ formed, so more $\ce{HC}l$ would have to be used. $\endgroup$ – SendersReagent Mar 27 '16 at 12:11
  • $\begingroup$ Related: Question regarding nitric acid, magnesium powder and initial reaction rate $\endgroup$ – Loong Mar 27 '16 at 14:02
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    $\begingroup$ A screenshot or picture of an exercise is not searchable. Please consider rewriting it (which I did already - please consider this for future reference), so that it can be of help for future visitors. $\endgroup$ – Martin - マーチン Mar 29 '16 at 7:10
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The reaction that occurs here is $$\ce{Mg (s) + 2 HCl (aq) -> MgCl2 (aq) + H_2 (g)^}.$$

As the reaction progresses, $\ce{H_2(g)}$ is released and hence the mass decreases.

I believe you are right, both I and II could be correct.

II, because a greater quantity of $\ce{HCl}$ will lead to more $\ce{H2}$ leaving, hence a greater mass decrease.

I, because putting the magnesium in smaller pieces is likely to increase the speed of the reaction, and hence increase the mass of $\ce{H2}$ that has been created in a given time. However, since the end of both functions is "flat", one could assume that the reactions won't evolve any more, thus invalidating this answer (if one waits long enough for the reaction to not evolve any more, the speed of the reaction will in most cases not affect the end result).

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  • $\begingroup$ The problem with $I$ is that while, yes, the rate would be slower, the final mass would be the same. Here, at time $t=\infty$ $m_{II}\ne m_I$. $\endgroup$ – SendersReagent Mar 27 '16 at 11:45
  • $\begingroup$ @DGS I didn't really consider that the end of the given timeline was an equilibrium state... but now that you're pointing that out, it indeed may be what the graph meant to convey. Amended. $\endgroup$ – Hippalectryon Mar 27 '16 at 11:54
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    $\begingroup$ Well, the reaction won't be in equilibrium due to evolution of $\ce{H2}$ (though, I don't think it would be in equilibrium either way). Since both lines go horizontal, both would be at completion, so they should have converged by end of the graph. $\endgroup$ – SendersReagent Mar 27 '16 at 11:59
  • $\begingroup$ The first word of the question is "Excess", which means that all of the dissolved HCl will be converted into aqueous magnesium chloride solution. Starting with the same amount of magnesium (and acid) the final mass must be equal. | @DGS wouldn't the rate be faster with smaller sized grains of magnesium? $\endgroup$ – Martin - マーチン Mar 29 '16 at 7:01
  • $\begingroup$ @Martin-マーチン Of course it would, but the final masses would have to be the same. Same amount or hydrogen ions to make hydrogen means same amount of gas evolved. These have gone to completion, or at least you can see the asymptote they're each trending toward. Clearly different. But if higher concentration acid was used, but same volume, that means more magnesium can react and more hydrogen gas can evolve, thereby ending with a lower final mass. $\endgroup$ – SendersReagent Mar 29 '16 at 7:11
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I. False – because the magnesium has a higher surface area in this reaction, the rate will proceed faster, producing a steeper slope in the graph; however, the final mass of the solution will be the same in both experiments. We would note that the lines converge when each system reaches equilibrium.

II. True – the higher concentration of $\ce{HCl}$ means more $\ce{H2}$ gas will be lost from the system at equilibrium. Thus, the line would flatten beneath line 1.

III. False – a lower temperature would mean the average kinetic energy of molecules in the system would be lower; therefore, a smaller fraction of the molecules will have sufficient energy to react with others. In this case, we would note an elongated curve that converges with the first.

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  • $\begingroup$ Please don't use MathJax as a styling element and also please don't use sizing commands where not necessary. $\endgroup$ – Martin - マーチン Mar 29 '16 at 7:14

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