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Concentration and equilibrium graph HSC question (1 Viewer)
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rand_althor
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As CH3OH is a product, its initial concentration would be zero as the reactants haven't reacted to form it yet. This means the bottom dashed line represents CH3OH. Over time as the CH3OH is being made, the middle dashed line decreases so that it has a value half that of the solid line. This indicates that the middle dashed line is the H2, as in the reaction it has a molar ratio of 2:1 with CO. So that leaves the solid line at the top to be CO. From this, you can see that at T1 H2 is added, and at T2 CH3OH is removed. So the answer is A.
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BlueGas
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Oh okay, I understand that CH3OH's initial concentration is 0, but the H2 and CO part confuses me, is there a simpler way to know which curve is for which?
rand_althor
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From the reaction, you can see that CO and H2 react in a 1:2 ratio. This means that for every 1 mole of CO, 2 moles of H2 is required to produce CH3OH. This also means that their molar concentrations are also in the ratio 1:2. Let's number the y-axis from 0M to 10M, with increments every 1M. Both CO and H2 start at an initial concentration of 10M. At time T1, the solid line has dropped to 7M, and the middle dashed line has dropped to 4M. The middle dashed line has dropped double the amount that the solid line has, i.e. the substance being represented by the middle dashed line has been used up twice as much as the substance being represented by the solid line. Since we know that CO and H2 react in a 1:2 ratio, this means that the solid line represents CO and the middle dashed line represents H2.
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Librah
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It doesn't get simpler than that.
A little late to the party bc I'm preparing for the HSC and accidentally found this question online.
When I do this question, I think the best way to approach it is to look at the "shape" of the graph. Ignoring the increase/decrease cliffs of the curves and you would realise that the solid line and the "densely dotted" line look very similar to each other whilst the bottom most line's shape is completely different from the rest. This suggests that the behaviour of the two similar lines is this means that both of these substances should belong to the same side of the equilibrium. Coincidentally, the side of the equilibrium with CO and H2 is the only side with 2 substances hence the two similar curves must belong to these two substances and the bottom most line will be representing the CH3OH.
As for distinguishing which line is for H2 and which line is for CO, look how much the gradient change when a disturbance is introduced to the equilibrium. The line with a steeper change in gradient means that more of that substance will be consumed during the rebalancing of the equilibrium (due to stoichiometry). In this case, one instance of reaction between CO and H2 will consume 1 molecule of CO and 2 molecules of H2, thus we say that H2 is consumed more rapidly that CO. From here, it is obvious that the densely dotted line falls much quicker than the solid line before hitting the first "cliff", hence the densely dotted line must be H2 and the solid line must be CO.
(Hope this doesn't confuse anyone even more)
Good luck with your HSC guys!