CHEMISTRY COMMUNITY

Janine Chan 2K

I think it's just that when doing the problems, we can assume A is the same and that they cancel out, but in real life A is actually different.

Janine Chan 2K

David Zhou 1L wrote:For this though, couldn't the Fe be oxidized into Fe²⁺? Why does it have to be Fe²⁺ oxidized into Fe³⁺?

I believe it's because we have FeCl2 as a reactant, which would imply the iron on the reactants side must be Fe2+.

Janine Chan 2K

The +1.23 volts is for the oxidation reaction. The standard reduction potential is actually -1.23 V so the solution manual is correct. And you want to get the most negative value of E because in an electrolytic cell, the reaction is non spontaneous. Thanks, I see what you mean. But if the +1.23 vol...

Janine Chan 2K

I was confused at this too, but I think it's asking about more about the salt bridge. The external circuit is where the flow of electrons happens, but we need electrolytes to complete the circuit. Since e- go from anode to cathode, electrolytes will then go from cathode to anode. https://www.chem.wi...

Janine Chan 2K

Did you mean standard enthalpy? We aren't able to directly do H_fproducts-H_freactants because the question gives values for heat of combustion (not formation).

Janine Chan 2K

The anode should be written on the left and the cathode should be on the right. In this reaction, Cl2 is being reduced, so the half reaction with Cl2 and Cl- is on the right.

Janine Chan 2K

The solution manual states that the cathode should have the least negative E and the anode should have the most negative E. However, it then says that the cathode, which is Ni2+, has a E of -0.23 V while the anode has a E of +1.23 V. Isn't this contradictory? Also, if the point is to make a cell wit...

Janine Chan 2K

"Write balanced half-reactions for the redox reaction of an acidified solution of potassium permanganate and iron(II) chloride." How do we know to use Fe³⁺ + e- --> Fe²⁺ as the half rxn? Why not Fe²⁺ + 2e- --> Fe or something else?

Janine Chan 2K

In 8.75c, 4 C-H bonds are still broken. It's just written as 1 in the answer key because it cancels with the 3 C-H bonds formed in CH3Cl in the products.

Janine Chan 2K

We can assume the rate law using the coefficients in an elementary reaction only when proposing a reaction mechanism.

Janine Chan 2K

Catalysts increase the rate of reaction by providing a new pathway for the reaction with a lower activation energy.

Janine Chan 2K

Nope you're good, if you divide both the initial and final mass by liters to find the concentrations of each, the liters will cancel out when you plug into [A]_t=[A]₀e^-ktanyway.

Janine Chan 2K

Since 15% isn't a multiple of 1/2, we have to do more calculations. We first solve for k using the 1st-order equation t 1/2 =0.693/k. If we end up with 15% of the initial concentration, we can say that [A] t =.15, and [A] 0 =1. By rearranging the equation ln[A] t = -kt + ln[A] 0 , we can solve for t...

Janine Chan 2K

So we have rate=k[A][B] 2 [C] 2 . The initial rate is given in units (mmolA)*L -1 *s -1 . Each of the concentrations are given in units mmol*L -1 . When we multiply all the concentrations of reactants together, we end up with units mmol*L -5 . By dividing (mmolA)*L -1 *s -1 by mmol 5 *L -5 to find u...

Janine Chan 2K

I think the reaction rate is always written as positive. In the solutions manual, it makes up for the decreasing concentration by adding a negative sign in the equation (-∆conc/∆time). So, you would just say that the reactant concentration is decreasing at (input rate here).

Janine Chan 2K

Are the coefficients of the reactants correlated to the order/rate law of a reaction? Or is the rate law only dependent on the concentrations and rate constant given? I guess in other words, can we determine the order/rate law solely by looking at an equation and the coefficients? For example, in 15...

Janine Chan 2K

The reaction rate = ∆concentration of reactants or products/∆time, which would give you units of molarity/time, which is usually mol*L -1 seconds -1 . If you look at the the general equation for the differential rate law, it is rate = k[R] -1 . For example, for a first order reaction, the rate = k[A...

Janine Chan 2K

So to calculate the reaction rate, we usually do reaction rate = ∆concentration of reactants or products/∆time. This would give you Molarity/Time, which is usually mol*L -1 seconds -1 . If you look at the the general equation for the differential rate law, it is rate = k[R] n . Part a of this questi...

Janine Chan 2K

Doesn't ln[A] vs. time have to be linear to assume it's first order? Or in this case we're just saying that in this question, we can see that multiplying by a factor of 1.2 will increase the rate by a factor of 1.2, which aligns with the first order differential rate law rate = k[A].

Janine Chan 2K

The values are given to you for pH=7 at the beginning of the problem set!

Janine Chan 2K

In this question, how do we know one of the half reactions should include water?

Janine Chan 2K

Where do we get the equation 3Tl+ --> 2Tl + Tl 3+ from? Wouldn't Tl+ e- --> Tl also be Tl+ "disproportionating" in solution?

Janine Chan 2K

For the anode, why is it written in the order Ag(s)|AgBr(s)|Br-(aq)? Is this because Br- is an ion/aqueous so it's written closest to the salt bridge? If solids are written on the outside, why is Ag(s) written before AgBr? Is this because it's written from reactants to products? But if it's written ...

Janine Chan 2K

How do we know I2 isn't conducting?

Janine Chan 2K

Degeneracy can't be 0, but entropy can be 0 if it's a perfect crystal (aka when W=1 because there is only one state for it to exist in). When W=1, ln(1)=0, making S equal to 0.

Janine Chan 2K

I believe the signs don't change due to the way the question is worded. We don't assign the interior or the exterior to be the "system," so we don't say the ∆S of the "surroundings" is -∆S of the "system." The question simply asks which is greater: the change in entropy...

Janine Chan 2K

Since entropy is a state function, the pathway taken from final to initial doesn't matter. ∆S_sys for irreversible and reversible should be the same.

Janine Chan 2K

In order for decomposition of compounds into their elements to be stable, ∆G of decomposition would have to be (+) aka it won't spontaneously break down. Since decomposition is the opposite of formation, ∆G f would have to be (-). In other words, the formation is spontaneous and the decomposition is...

Janine Chan 2K

In the solution manual error doc, a different value of C_v is used for each gas. Why is this?

Janine Chan 2K

Yes, we are supposed to use the equation ∆G = ∆H_vap - T∆S_vap. We use the vaporation values because NH3 is going from a liquid to a gas.

Janine Chan 2K

The solution manual error says Cv but uses 5/2 R, which is the value for Cp. Maybe the doc has a typo too?

Janine Chan 2K

There are 6 different states the molecule can exist in (you can switch the Os and the Fs around the S atom 6 different ways).

Janine Chan 2K

In the solution manual errors PDF, the equation ∆S=nC_vln(T₂/T₁) is used, with C_v being 5/2 R

Janine Chan 2K

In 9.13, why is Cv used instead of Cp when finding the change in entropy resulting from the increase in temperature? Cv is the heat capacity at constant volume, and volume in this question isn't constant

Janine Chan 2K

Also, the more microstates there are, the higher the entropy.

Janine Chan 2K

I believe it stands for standard molar entropy, the entropy of 1 mol of substance at standard state

Janine Chan 2K

In general, more complex molecules will have higher entropy. Since HBr is larger than HF, it is more complex, has more particles (and thus more possible states it can exist in), and higher entropy.

Janine Chan 2K

The way I thought about it was that when we first look at the equation, there are 2 mols on the left and 2 mols on the right. However, the reactants side has a mol of gas, which has more entropy than liquids are solids. Aqueous just means it's dissolved in water. A gas would have more entropy than a...

Janine Chan 2K

When writing the chemical equation, how do you know water is in the liquid state?

Janine Chan 2K

Also, in reversible expansion, the external pressure always matches the internal pressure. If there is any increase in pressure, external and internal energy will match each other. So, reversible reactions are always doing the maximum possible amount of work. In a reversible reaction, the process is...

Janine Chan 2K

In lecture, we discussed how in a perfect system, q system + q surroundings = 0. If this is the case, q system = - q surroundings . In this sense, you could say it's true that the negative sign can go on either one, because q system = - q surr is the same as q surroundings = - q system . If you assi...

Janine Chan 2K

In the textbook, the question states that it takes 10.5 hours to reach 5C, but the solution manual uses 10 h. Is this a typo or am I missing something? I used these equations (4.184)(150 g)(5C - 0C)= 3135 J for 150 g and then I did ( \frac{3135 J}{150 g}*\frac{18.02 g}{1 mol}*\frac{1 kJ}{1000 J}*\fr...

Janine Chan 2K

1.30*10⁵ J is the amount of energy it takes to raise the temperature of water which is calculated by (400.0 g)(4.18)(100.0-22.0). 1.45*10⁵ is the total amount of energy it takes to raise the temperature of both the water and the kettle itself.

Janine Chan 2K

When going from a solid to liquid, aka melting, heat is required without a change in temperature because it is a phase change. So, q is positive. For work, say ice is the solid. Ice, the system, gains energy, while the surroundings lose energy. This implies work is negative (because the system loses...

Janine Chan 2K

The kettle should be part of the system, which is why the kettle is included when calculating the heat that must be supplied. I guess if you think about it, the kettle must be heated first before the water can be heated.

Janine Chan 2K

Here is a mathematical approach in addition to the responses above... At constant volume, the heat transfer is interpreted as ∆U. This is true because ∆U = q + w, and w = -P ex ∆V, where P is external pressure and V is volume. If the volume is constant, ∆V=0, which implies w=0, and thus ∆U=q. At con...

Janine Chan 2K

The terms energy and heat are also related in the equation ∆U = q + w, with ∆U being the change in internal energy, q being heat, and w being work. In other words, heat and work are both means of transferring energy.

Janine Chan 2K

Also, reversible reactions occur slowly, while irreversible reactions occur quickly. So, if the process is reversible (done slowly), minimal energy is lost to the surrounding as heat, allowing for maximum expansion work.

Janine Chan 2K

Reversible processes are very slow. Because it is done slowly, less energy is lost to the surrounding as heat. In the gas and piston example, an extremely small increase in external pressure would cause the gas to be compressed and the piston to move in. If the process were to be done quickly (aka i...

Janine Chan 2K

The example given in class compared HClO with HBrO and HIO and said that HClO is more acidic. Since Cl is more electronegative, it takes e- density away from O, thus stabilizing it (a stable resulting anion would drive the reaction forward). Taking e- density away from O makes the compound a stronge...

Janine Chan 2K

We multiply 200 ml by .025 M to find the number of moles that were actually made. Then divide by 250 ml to get the molarity of the actual solution.

Janine Chan 2K

Reactions that include bond breaking are generally endothermic. EX: Cl₂ <--> 2Cl. Combustion reactions are almost always exothermic.

Janine Chan 2K

Yes, I believe if it just asks you to name the compound and there is no charge, you don't need to write "ion." If it's a complex ion (aka has a + or - charge), you'll need to write ion at the end of the name.

Janine Chan 2K

Pi bonds in benzene would be delocalized. When you draw the Lewis structure, it alternates between single and double bonds. In reality, there are no alternating bonds, which is why benzene is usually drawn with a circle inside a hexagon. The p-orbitals overlap side-to-side and electrons are delocali...

Janine Chan 2K

If you're given moles and liters, you'll have to convert to molarity. K can only be found with concentrations (molarity) or partial pressure (where the units will be something like atm, bar).

Janine Chan 2K

The s-character in sp2 is more than that in sp3. The way I see it is in sp2, the ratio of s:p is 1:2, while the ratio in sp3 is 1:3. If the bond angle in sp2 is larger, that would mean an increased s-character would result in a larger bond angle.

Janine Chan 2K

When the reaction lies to left, there are more reactants than products. If the equilibrium is shifted to the left, there are more products than reactants. Equilibrium will shift to the side with fewer moles, so if there are products, equilibrium will shift to the reactant side (left side).

Janine Chan 2K

I believe the lines and triangles you are referring too are just bonds between two atoms. The lines and triangles symbolize depth. If you imagine your paper to be the plane you're drawing the VSEPR structure on, the lines fading out would indicate an atom that is going into your paper, and the trian...

Janine Chan 2K

When molecules have lone pairs around the central atom, other bond angles are forced closer together (we won't know the exact angle, but you can see that the bond angles would be a little less than those in a structure without lone pairs). This applies to O3, which has 1 lone pair around the central...

Janine Chan 2K

There actually is no lone pair around the S. I believe the Lewis structure is just a sulfur atom in the middle, surrounded by 3 oxygen atoms and 1 other sulfur atom. Since there are 4 bonding pairs around the central atom, it would be a tetrahedral shape.

Janine Chan 2K

In a trigonal planar shape, there are only 3 bonding pairs surrounding the central atom. As you drew in your Lewis structure, ICl3 has 3 bonding pairs and 2 lone pairs, which would make it T-shaped.

Janine Chan 2K

Nitrogen also has a half-filled sub shell, which means it has lower energy and is more stable. Since it would be more difficult to take away an electron from a more stable subshell (aka higher ionization energy), Oxygen has the lower ionization energy.

Janine Chan 2K

It should be (1). Higher energy electrons are removed first. In this case, the 4s electrons have a higher energy than 3d.

Janine Chan 2K

I believe in this case "penetrate" just means how close the electron can get to the atom. So since it "penetrates" more, it is more attracted to the nucleus and therefore more effective at shielding other electrons than orbitals that are further out from the nucleus.

Janine Chan 2K

^ The electron configuration of Helium is 1s^2, so yes, it is part of the s-block.

Janine Chan 2K

I think the only thing you're missing is that 41 should be in m * s^-1, not just meters.

Janine Chan 2K

I also had a question on this^
The solution manual uses a different character for Planck's constant, and it uses the equation ∆p∆x= h/2, whereas in lecture we learned the equation ∆p∆x=h/4pi.

Janine Chan 2K

So I used the equation was c/wavelength=R(1/n_1^2 - 1/n_2^2). Essentially, you're using the given wavelength and using c=v*wavelength to find the frequency, and then using v=R((1/n_1^2 - 1/n_2^2) to find n_2. (We also already know n_1 = 1 because a wavelength of 102.6nm is Lyman series

Janine Chan 2K

The formula basically relates the wavelength to the initial and final states of an electron. We can find the inverse of the wavelength by multiplying Rydberg's constant by the change in energy of the electron.

Janine Chan 2K

Leading zeroes are not significant (in 0.032, the zeroes before 3 aren't significant so there would be 2sf). Trailing zeroes ARE significant (in 0.03000, all the zeroes after 3 are significant so there would be 4sf).

Janine Chan 2K

If the compound includes carbon atoms, C will be written first. If it includes hydrogen atoms, H will come second. Any other elements should follow in alphabetical order. Hope that helps!

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