Skip to Main Content

Gen Chem 1 Passage Sample Questions


This is MCAT General Chemistry 1. We'll be doing some a few sample problems like those you would see on the actual MCAT. This is another passage based problem set. So again, we're gonna read through, but we don't try and read to enjoy ourselves from reading solely for information.

Passage II. Questions 7-11. Silane is a pyrophoric gas that is it spontaneously combust in the presence of air to produce silicon dioxide (an inert solid) and water. This happens via two-step reaction. 1 SiH4 + O2 goes to SiO2 + 2H2.

2 H2 + one-half O2 goes to H2O. Now when we see these two reactions we don't need to memorize them or study them in too much depth. We just want to know that they exist. There's two reactions in a two step process. Chlorinated silanes are not pyrophoric and are consequently more stable.

In closed and isolated containers, the various species of chlorosilane readily interconvert according to the following equilibrium reactions. Three, four, five, we have different equilibrium reactions. We have this silane with the chlorine, two chlorines, three chlorines, and they just go back and forth to each other. With monochlorocailane is slowly fed into a pre-filled cylinder of pure oxygen, eventually the monochlorocailane will reach a threshold concentration required to initiate reaction three.

Under these conditions, any silane created is immediate combusted. Thus, the net equilibrium reaction is 6: 2SIH3Cl(g) + 2O2(g) goes to SIO2(s) + 2H2O(l) + SIH2Cl2(g). Equilibrium for this reaction is heavily skewed in favor of the products. Heats of reaction are given for the six reactions in the following table. And then we have a table that has reaction and then the entropy associated with that reaction.

We don't memorize everything that's in the tables, we just wanna know what's in them, and where to find them if we need them. Numer 7. In the pyrophoric combustion of silane, hydrogen gas is a(n). We can walk through these answers, a transition state, an accelerant, an intermediate, or a byproduct.

By byproduct is generally something you end up with at the end, and hydrogen is fully consumed by the end of two. So you say probably not byproduct. An intermediate is something that is created in one reaction and then consumed in the next. That seems to be the case for hydrogen, so we're going to put a little check mark there.

An accelerant would be something that would speed up this reaction. That doesn't seem to be the case. And a transition state is something that exists for a very brief second of time. It really couldn't exist on its own, so most likely, it's not a transition state either, and we're going to go with intermediate. This was purely a definitional question.

Eight, the compound with the largest dipole moment is, for this we want to remember that a dipole moment results when you have elements that have different electronegativities in a bond. Now the geometry of the molecule will also affect the dipole moment. So we got, for this case, silicon tetrachloride. Now silicon is similar to carbon.

It's the element directly below carbon in the periodic table. So we would assume this would form a tetrahedral shape, When it has four atoms on it. So we have a Cl, Cl, Cl, and a Cl. Tetrahedral, remember the bond angle and the tetrahedral shape is 109.5.

So now the question is, is the Si-Cl bond polar? Well it is, however since we have four of them all pulling against each other, symmetry will cancel out any of these polar bonds for the tetra chloride. So there will be no dipole moment on SiCl4. Silane, SiH4 will be similar. The SIH bond even if it is polar, because there are four of them, they would cancel each other out.

So then it leaves us with the monochlorosilane, and the dichlorosilane. We draw these. The monochlorosilane has one chlorine, The dichlorosilane, Has the two. Now this will have dipole moment over here. The dichlorosilane has two polar bonds.

And the net result of this two is we get a slight cancellation between these two. So overall it reduces the dipole moment of the dichlorosilane, so the monochlorosilane where there is no competing effects of the other bond actually has the largest dipole moment. Symmetry cancels out a portion of the dipole moment for the dichlorosilane. Number 9.

According to Reaction 4, which of the following changes will increase the amount of trichlorosilane in the system? Anytime you see phrasing like according to Reaction four, that's a dead give away that you should actually look at Reaction four. Now this problem is a LeChatelier's principle type problem. We wanna know what sort of pressure we can exert on this system to increase the amount of the trichlorosilane which is this SiHCl3.

First up A, adding a catalyst to the system. Remember catalysts do not affect the amount, just the rate of conversion backwards and forwards. So we just go back and forth between the products at a faster rate. We don't change the amounts of products reactants. So A would not increase the amount B says increasing the system pressure.

Well, let's look over here. If had 1 mol of a gas on this side. Here we have half a mol and another half a mol, or essentially 1 mol. So since changing the pressure, increasing it wouldn't affect the amount of mol of gas in the system. This will do nothing to change the amount of trichlorosilane.

Changing in the pressure doesn't change the moles of gas. Increasing system volume, pressure volume are related in effect that if we increase the pressure were decreasing volume or if we increase the volume or decreasing the pressure, we saw there was no effect for pressure change. There will also be no effect for a volume change. So by default, we're left with cooling the system temperature.

But is there a way we can check this? Over here we have nothing that tells us about system temperature. However, up in the table, we can see that Reaction 4 is Exothermic, meaning it gives off energy. If something is Exothermic and we cool it down, it will release heat to make up that difference in order to release the heat.

It's gonna have to shift right, that would increase the amount of Trichlorinated Silanes. So cooling the system will increase the amount of trichlorosilane as the system breaks up the dichlorosilane to produce heat with the trichlorosilane. So the answer here is D. Number 10, the equilibrium constant for reaction six is, now on a problem like this often it's best to write out what you think the equilibrium constant will be, and then compare it to the available answers.

So an equilibrium constant KEQ is equal to concentration of the products, over the concentration of the reactants. We can then just fill this in, we're looking at reaction 6. Our products will give us SiO2, H2O, and the dichlorosilane SiH2Cl2 over reactants are and .

Now remember in an equilibrium constant these stoichiometric coefficients become exponents. So we're going to have H20 to the second power. SiH3Cl to the second power and 02 to the second power. Now we also remember equilibrium constants do not contain solids and liquids. These drop out.

So Si02 is out and H20 is also gone. This reduces our equilibrium constant to SiH2Cl2 over SiH3Cl squared times O2 squared. We can now look through the answer choices and see if we can find a match. SiO2, well we know that one's wrong because it's got the solids in it. This one has a liquid, H2O in it.

Two possibilities, SiH2Cl2 over SiH3Cl2 squared times O2 squared or SiH2Cl2 over 2 SiH3Cl2 times 2 O2 squared. The only one that matches our equilibrium constant is B. In C, they've accidentally included the 2 that was a coefficient as a multiplier. We do not do that. The last question, number 11.

What is the net heat of combustion for silane in the two-step pyrophoric reaction? Well since we have enthalpies for the individual reactions, we can use Hess's Law to add these, silane + oxygen going to silicone dioxide + 2H2. H2 + half O2 goes to H2O. Delta H for reaction 1 is = to -1262 joules.

Delta h for reaction two is equal to 137 Joules. Now to get the net reaction, we're gonna want to get the hydrogen to eliminate. So to do this, we're gonna need twice as many hydrogens, twice as many of these oxygens, and twice as many of the water. This also multiplies our delta h. So then you'll have 2H2 cancelling with H2.

Two of this will be 1 O2, and then 2 H2O. So the net reaction is silane + 2 O2 yields silicon dioxide + 2 H2O. We can check to make sure everything's balanced. 1 silicon, 1 silicon, 4 hydrogens, 4 hydrogens, 4 oxygens, 2 + 2 or 4 oxygens.

So our elements are all balanced. Now delta H is going to equal -1262 + 2(137). Or -1262 + 274 = -988 joules, and that's the choice D. So, again remember if you have a delta H, a delta G, or a delta S, you need to multiply these by the stoichiometric coefficient.

And that concludes the set of problems.

Read full transcript