<p>HOI is weaker than HOCl because the chlorine atom has a higher electronegativity than the iodine atom. Because of this, Cl is able to hold more closely to the electrons of oxygen, creating a high electron density in the O-Cl. This weaken the O-H bond, allowing H to ionize more easily. Since we measure the strength of an acid by how well the H can ionize, clearly HOCl is the stronger acid.</p>
<p>For 6 c), I’m pretty sure the answer has to be p-orbitals. I know that it hybridizes to sp3 and thus has angles of ~109.5, but given that it says the angle are between 90 degrees, AND especially important is that they say what ATOMIC orbitals, not hybrid orbitals, it must be p-orbitals that are involved in the bonding. All p-orbitals are mutually perpendicular, each located on a different axis(x,y,z), so they would all be 90 degrees.</p>
<p>Also, I’m pretty sure that 5 e) is false. Gibbs free energy merely predicts spontaneity. The more positive the free energy, the lower the concentration of products with respect to the reactants ONCE EQUILIBRIUM IS REACHED. However, Gibbs free energy does nothing to predict the RATE of reaction. Thus, a reaction that is less spontaneous than another can occur faster.</p>
<p>5 a) is right, except for the fact that I think one must use P(molecule) so that you are doing the equilibrium pressure ratios, and not [molecule] which would imply concentration, which is Kc</p>
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<p>you do know that this statement will earn you no points, since you have to explain the trend.</p>
<p>Aww I miss AP Chem, I took it last year.
But HOCl is a stronger acid b/c the Cl makes it more polar and the H’s bond weaker. HOCl4 is even stronger.</p>
<p>what explains H2S’s close-to-90 degree bond angle whereas H2O is 100+</p>
<p>I really don’t think that H2S has a 90 degree bond angle haha…I believe its bond angle is ~105ish, like H2O. However, given that they tell you that it has 90 degree bong angle with its atomic orbitals, then you have to ignore the sp3 hybridization and just say its the p-orbitals.</p>
<p>Edit: I just read up on it. H2S is hybridized like H2O, but only very slightly. Its bond angle is around 92 degrees. This is in part because S is larger than O, and the electrons are already far enough apart that they don’t have to hybridize as much to lessen the repulsion between electrons. Thus, it is almost exclusively atomic orbitals.</p>
<p>[Hydrogen</a> sulfide - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/Hydrogen_sulfide]Hydrogen”>Hydrogen sulfide - Wikipedia)</p>
<p>[Water</a> (molecule) - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/H2o]Water”>Water - Wikipedia)</p>
<p>Sorry…I just edited my last post haha…H2S definitely is about 90 degrees.</p>
<p>5e) Yeah, this is false. Several industrial processes have alot of negative free energy potential, but are extremely slow. This is why they use a catalyst to speed up the reaction rate. Gibbs free energy only tells you about the spontaneity of a reaction, but nothing about its rate.</p>
<p>Hm, anyone know what the answer to 3e was? (identifying the order of CH4 and CL2 in the given mechanisms.). Wasn’t sure about this one. Also unsure about the explanations for questions 4a, b, and c.</p>
<p>Edit: never mind, i give up.</p>
<p>I’m pretty sure CH4 is first order, and Cl2 is 1/2 order…what do you mean the explanations for 4a, b, and c? The secondary questions or why those are the equations??</p>
<p>For 5(d), the total bond energy of the reactants is greater than that of the products, even though the reaction is endothermic. An alternate way of finding the enthalpy change is to subtract the total bond energy of the products from the total bond energy of the reactants, so if the enthalpy change is positive the reactants must have more total bond energy. I think this formula was on the formula sheet.</p>
<p>5d I put reactants as well.
5e I also put false, because rate of reaction doesn’t have anything to do with Gibbs
4e I guessed but I put first order, zero order</p>
<p>If I missed 3b but did the right calculation for 3c based on my answer on 3b, I would get credit right? I stupidly forgot to divide by 6.02e23 in 3b.</p>
<p>On the kinetics question the proposed mechanism has the fast equilibrium step first so it’s a different process then it would be if the slow rate-determining step was first. If the rate determining step was first, the rate law of the slow step would be the same rate law for the overall action. But that’s not the case…</p>
<p>Since the fast equilibrium step is first, it forms a bottleneck and you have to write that Rate(fwd)=Rate(rev) which would be k[Cl2]=k-1[Cl]^2. Then you write the rate law for the slow step which is k2[CH4][Cl]. Since an intermediate can NOT be in the overall rate law, you solve for the intermediate from the equated fwd and rev rate laws for the fast equilibrium. Therefore [Cl]=sqrt(k[Cl2]/k-1). Then you plug this quantity in to substitute for [Cl] in the slow step and now the overall rate law is K’[CH4][Cl2]^1/2</p>
<p>Hey Klebian, thats exactly what I did. I also forgot to divide by 6.022*10^23 so my answer for the wavelength part was wrong. I think you will miss the first answer but completely get the second answer right because they dont have double jeporidy. Otherwise this would be unfair…as for number 2, if you missed part a, you would miss eveyr single part of the problem invovling calcs because they all use a. Same with other problems but with smaller parts.</p>
<p>apparently, the curve for recent years has been around 107 or 108 out of 160.</p>
<p>have curves even earlier been lower or higher than 107?
I mean for history, the generic curve as 114/180, and for AB calc, the curve was around 75/108, but for many practice tests, the curve tended to be lower than the “generic curve.”</p>
<p>anyone have an idea for ap chem?</p>
<p>For 5e, I believe it is no.
My teacher taught us that the rate of a reaction can never be determined by the spontaneity of a reaction or thermodynamics.</p>
<p>Can anyone post what they asked for 3. (b)?</p>
<p>@HiPeople
Actually, our ap chem teacher said that if you get the first calculation wrong, but your work for the following answers are in accordance with the first answer with the correct steps, then you will get points.</p>