CHEMISTRY COMMUNITY

Niharika Reddy 1D

Does leaving out the hydrogens apply for drawing chair and boat conformations as well? In past final questions, all the chair conformations have hydrogens attached or just have the axial/equatorial up/down lines drawn in, but I was wondering if it would be okay to leave them out. I think the "1...

Niharika Reddy 1D

Adding a solid would change the cell voltage if the concentrations of ions are changed. Can you post the whole question?

Niharika Reddy 1D

The double bond does get priority but the correct way to number it also gives a lower number for the first substituent than how you numbered it (1<2) while still having the double bond at carbon 1. Usually, it's better to start numbering on the carbon that has a substituent attached AND has a double...

Niharika Reddy 1D

We can usually look at the moles of gas to predict the sign of ΔS. If more moles of gas are produced than were initially present as reactants, there is an increase in entropy and ΔS is positive. In this question, since the reactant side has 3 moles of gas whereas the product side has only 2, there's...

Niharika Reddy 1D

If you balance the reaction for the combustion of 1 mol octane, you will get 8 mol CO₂. Dividing ΔH_c° of octane by 8 mol CO₂ yields -684 kJ/mol CO₂.

Niharika Reddy 1D

Heat capacity increases with molecular complexity. As more atoms are present in a molecule, there are more possible bond vibrations that can absorb added energy. In addition to having more atoms, ethane has a C-C single bond, which allows it to have rotational freedom, whereas ethene has a C-C doubl...

Niharika Reddy 1D

On the C that is attached to the carboxyl group and chlorine, Cl gets priority. We look at just the C in the COOH group, not the whole group collectively when determining priority, so since Cl has a greater atomic number than C, Cl receives priority. Both groups with priority are on the same side of...

Niharika Reddy 1D

Heat capacity increases with molecular complexity. As more atoms are present in a molecule, there are more possible bond vibrations that can absorb added energy. In addition to having more atoms, ethane has a C-C single bond, which allows it to have rotational freedom, whereas ethene has a C-C doubl...

Niharika Reddy 1D

For this particular question, I don't think the answer is affected by looking at the second Cu half reaction you listed, since both of those reduction potentials are much larger than those of the other metals. When I looked up the half reactions and their standard potentials, I looked at all the hal...

Niharika Reddy 1D

Yes, functional groups get highest priority when numbering, but this alkene does not have any typical functional groups attached (atoms or atom groups that involve other elements in addition to carbon and hydrogen atoms). In the intro to ochem book, it says carbon-carbon double and triple bonds may ...

Niharika Reddy 1D

I believe so, as radicals have an unpaired electron and are highly reactive and unstable. They seek to gain electrons, which is why they're electrophiles.

Niharika Reddy 1D

Your association between size and polarizablility is correct (larger=more polarizable) However, I- is larger than Cl-, not smaller. As you go down a group in the periodic table, the atomic radius increases as shielding increases and the increased number of electrons are more loosely bound to the nuc...

Niharika Reddy 1D

I'm not sure if you mean the rate law for a particular elementary step by "elementary rate law", but intermediates should not be included in the final rate law for the overall reaction after taking all the different elementary steps of the reaction mechanism into account. If an intermediat...

Niharika Reddy 1D

When a question does not specifically state what the order of the reaction is, but gives you the rate constant (k), the order of the reaction can be determined from the units of the given rate constant. k is 8.39 M -1 .s -1 , so the reaction must be second order. This is how to calculate the units f...

Niharika Reddy 1D

: Half life for second order

At the half life, t_1/2, half the initial concentration remains. Substituting this piece of information into the second order integrated rate law and solving for t gives the equation for the half life of a second order reaction.

Niharika Reddy 1D

We subtract them to see how much of the initial has been used up and what concentration is left after producing that much of the product B.

Niharika Reddy 1D

I'm not sure but I think you are referring to problem 17. When finding order of the reactants from data of experiments with their initial concentrations and initial rates, you should find 2 experiments in which the concentration of one reactant changes but the concentrations of the other reactants s...

Niharika Reddy 1D

Radical chain initiating reaction means the reaction yields a product that is a radical- one that has an unpaired electron. Since Cl has 7 valence electrons and NO has 11, both of these products are radicals because they have an unpaired electron on the Cl and N respectively, which can be seen from ...

Niharika Reddy 1D

The solutions manual is correct. The reason the concentrations are switched is because the equation was rearranged so the negative sign was "dropped". A property of logs is you can bring the exponent of the log down and out to the front, and vice versa. They put the negative sign back up a...

Niharika Reddy 1D

We have solid ice at 0.0°C. To raise the temperature of the ice, we must first melt it and then raise the temperature of what is then water , after being converted from ice. This is why the specific heat of water is used, since the ice has been melted into water. You would use the specific heat of i...

Niharika Reddy 1D

The textbook uses the equation E° cell =E° cathode - E° anode . When using this equation, the minus sign accounts for flipping the sign of the anode's reduction potential. Notice only reduction potentials are used when using this equation. If you flip the sign of the anode, which is the method I per...

Niharika Reddy 1D

Potential difference (E, or max potential) is measured as work(in J)/charge(in C). I don't think the q Sarah used in E=w/q is referring to q as in heat. I would say since work is a path function, potential difference, which is calculated using work, is a path function as well.

Niharika Reddy 1D

The initial and final pressures are the initial and final states. Since the gas is expanding under 2 different conditions but from the same initial pressure (1000. kPa) to the same final pressure (100. kPa), we can use the fact that entropy and internal energy are state functions to support the stat...

Niharika Reddy 1D

I think positional disorder relates to how much space the molecules have to move around. Gas in a larger volume would have more positional disorder than gas in a smaller volume, and molecules in a solid would have less position disorder than those in a liquid, which would have less than those in gas...

Niharika Reddy 1D

ΔS is a state function, so the entropy of the system will be the same regardless of the path taken to get to the same final state from the same initial state. This is why ΔS_sys is the same for both processes.

Niharika Reddy 1D

I was also wondering why the OH- was omitted from the cathode side when it is present in the reduction half reaction. If anyone has some insight as to why this is, it would be much appreciated! Thank you!

Niharika Reddy 1D

We initially have O 2 →H + We first balance the 2 oxygens by adding 2H 2 O: O 2 →H + + 2H 2 O Now, balance the hydrogens by adding H + . There are 5 hydrogens on the right and none on the left, so add 5H + to the left. 5H + + O 2 →H + +2H 2 O Since H + appears on both sides, we can simplify by cance...

Niharika Reddy 1D

As a side note, make sure you include states in the cell diagram!

Niharika Reddy 1D

I think the salt bridge is always included in the cell diagram as || to separate the anode on the left from the cathode on the right. Since a proper cell should always have a salt bridge to function correctly and avoid charge buildup, the salt bridge is an essential part of the cell diagram, which o...

Niharika Reddy 1D

My TA said that for ΔU, we would use C=3/2 R since we are at constant volume and plug that into q=nCΔT. Since constant volume means w=-PΔV=0, it follows that ΔU=q+w=q+0=q.

In this sense, ΔU and ΔH will not be the same, since C_v is used for calculating ΔU and C_p for ΔH.

Niharika Reddy 1D

I had the same question, and it was answered on a previous post: The exact reasons for this are quite technical and relate to the concept of standard states. In the end, if you want to use a "mixed" equilibrium expression, containing both concentrations and pressures, then the pressures mu...

Niharika Reddy 1D

Fundamentals Section K of the textbook (pages F78-F81) explains how to assign oxidation numbers with some examples. I have attached a picture of the rules for assigning oxidation numbers.

Niharika Reddy 1D

As you said, the oxidizing agent (OA) is itself reduced, and the reducing agent (RA) is itself oxidized. The OA causes the oxidation of the other species since it wants the electrons the species it oxidizes will give off, which it will then gain, causing the OA itself to be reduced. After you determ...

Niharika Reddy 1D

Since the equation given is already balanced (the total charge and number of each element are the same on both sides), either half reaction you look at will give you the same number of moles of electrons transferred (n). To split it up into half reactions, you can assign oxidation states to see wher...

Niharika Reddy 1D

A platinum electrode is used when there is no other suitable solid metal that is present in the half reaction for the side you are looking at (anode/cathode). Electrodes are metallic conductors, so without a solid metal electrode, the electrons won't transfer. Platinum doesn't interfere with the rea...

Niharika Reddy 1D

The half reaction for MnO 4 - in acidic solution is the following and it occurs at the cathode since this half reaction has the larger of the 2 half reactions' standard reduction potentials, and we want to devise a (spontaneous) galvanic cell: MnO 4 - + 8H + + 5e- -> Mn 2+ + 2H 2 O Since H + is pres...

Niharika Reddy 1D

Yes, you change the sign of E° when you flip the reduction half reaction, but if you multiply the half reaction by a constant, you don't multiply E° by the same constant as we did for enthalpy, for example, since E° is an intensive property and is always the same. If you use the equation E° cell =E°...

Niharika Reddy 1D

n is the moles of electrons transferred. If you write the balanced redox reaction correctly, both half reactions will have the same number of moles of electrons gained or given off, so it shouldn't matter which one you use.

Niharika Reddy 1D

This is what chem mod replied to the same question asked previously: "A common reaction is the decomposition of sodium azide, NaN3, to nitrogen gas and sodium metal." The term sodium metal means Na not Na+. All the metal elements with zero oxidation state are referred to as "metals&qu...

Niharika Reddy 1D

If K>1, the forward reaction is favored and the reaction will be spontaneous as given . This is indicated by a positive E° value. The question asks to identify which of the reactions have K>1, so we look at the given reaction to determine which half reaction occurs at the cathode/anode, not which or...

Niharika Reddy 1D

I got what you got as well. I think it may just be a typo in the corrected solutions, as it should be 1x10^2, not 1x10^12.

Niharika Reddy 1D

n is the moles of electrons transferred, so you're right in getting 1. This problem is on the solutions manual errors document, but I believe the final answer on the corrected solutions is incorrect, although the work leading up to it is right.

Niharika Reddy 1D

Why is it that partial pressure and concentration can be combined when calculating Q? For instance, for 13.41 a and c (I have attached a picture of the problem below), the solutions combine the given partial pressures of the gases with the concentrations of aqueous species to find Q, and multiply Q ...

Niharika Reddy 1D

No H+ should be left when it's in a basic solution, which is why OH- is added to both sides to cancel out the H+. On the side with H+, the OH- combines with H+ to form water, but on the other side, it remains OH-, which is acceptable since it's a basic solution.

Niharika Reddy 1D

I have attached a picture of my work for this question.

Niharika Reddy 1D

I think the way you approached it is valid as well, but since they gave the reaction quotient Q in the problem, they probably wanted us to find K to compare the two values and determine the direction the reaction will have a tendency to go in from there. As you said, both methods yield the same answ...

Niharika Reddy 1D

I got the same answers as Kayla.

Also, for part c of this problem, the solutions manual shows that P_Br(g)²=3.6x10^-15 bar, but this is the answer for just the partial pressure, P_Br(g), not the partial pressure squared.

Niharika Reddy 1D

There are C-H bonds in the reaction, but the same number of C-H bonds are present on both sides of the equation. If you were to account for them by saying 3 C-H bonds were broken, you would have to say 3 C-H bonds were formed, which cancels the bond enthalpy of the C-H bonds out when calculating the...

Niharika Reddy 1D

It depends on the type of heat capacity you are using. If it's specific heat capacity and the units are in J/g°C or J/g•K, use grams and your second equation. If it's molar heat capacity and the units are J/mol°C or J/mol•K, use moles and your first equation. Our objective is to find heat in J, so w...

Niharika Reddy 1D

Unless stated otherwise (as in problem 115 where it says the products of the combustion reaction are gaseous), the products of the combustion of an organic compound are carbon dioxide gas and liquid water, and any nitrogen present is released as nitrogen gas (N 2 ). If any other products are formed,...

Niharika Reddy 1D

No problem, glad I was able to help! :)

Niharika Reddy 1D

Yes, that's the reasoning behind it! Since volume is constant, there's no change in volume, thus work=0 and change in internal energy=q+w=q+0=q=+2.50kJ

Niharika Reddy 1D

I believe it's C instead of B because of the slopes of the heating curve at the different phases. The heat capacity of the liquid is greater than the heat capacities of the solid and gas (double to be exact), so the slope of different states' segments of the heating curve should reflect their heat c...

Niharika Reddy 1D

To solve for ΔS° r , or the standard reaction entropy, you use the standard molar entropies of the products and reactants in the balanced chemical equation. ΔS° r =(sum of nS° m (products))-(sum of nS° m (reactants)) Where n is the stoichiometric coefficient of the product/reactant from the balanced...

Niharika Reddy 1D

I think Martha may've switched the order of the initial and final volumes around in the equation for the work of a reversible process. It should be V₂/V₁, or V_final/V_initial; w_reversible=-nRTln(V_final/V_initial)

Niharika Reddy 1D

Yes, that's another way of saying it, but it's the same as what was stated above. Subtracting standard enthalpies of formation of reactants from the standard enthalpies of formation of products is the same thing as subtracting the standard enthalpy of the products by the standard enthalpy of the rea...

Niharika Reddy 1D

6 mol of C-H bonds are already present on the reactants' side, and we want to form 6 mol of C-H bonds in the end. The reason they're left out is because if you included them, they would end up canceling out since the positive value is used for the reactants, where bonds are broken, and the negative ...

Niharika Reddy 1D

We need to find the standard enthalpy of formation of dinitrogen pentoxide. The compound should be formed from its elements in their most stable states. For N 2 O 5 , the elements are nitrogen and oxygen, both of which are present as diatomic gas molecules in their most stable form. Hence, the react...

Niharika Reddy 1D

We can look those up in the textbook. I asked about that one too because it wasn't given in the workbook but was told we will be given any information like that that we would need on the real quiz, and we can just look it up in Appendix 2A for the practice quiz.

Niharika Reddy 1D

Those are standard potentials, but I don't think we use them until we get to electrochemistry. Which problem are you looking at?

Niharika Reddy 1D

Properties of logs allow you to bring the exponent (Avogadro's number in this case) to the front and multiply it after taking the natural log of the base (6). Since the equation is S=klnW, and W=6^(6.022x10^23=N A ), you can bring N A down to the front of the expression. This gives S=N A kln6=14.9J/...

Niharika Reddy 1D

Question 8.105 states: Vehicle air bags protect passengers by allowing a chemical reaction to occur that generates gas rapidly. Such a reaction must be both spontaneous and and explosively fast. A common reaction is the decomposition of sodium azide, NaN 3 , to nitrogen gas and sodium metal . For th...

Niharika Reddy 1D

I believe the "per mole" refers to the stoichiometric coefficients of the balanced reaction. So the delta H for that reaction would be ___kJ per 3 mol Al (s) for example, or per 1 mol NO (g).

Niharika Reddy 1D

The conversion factor (760 torr=1 atm) would be given to us. Although the back of the periodic table doesn't explicitly state that conversion factor, it gives the gas constant in L·atm/K·mol (.08206) and in L·torr/K·mol (62.364). Since we want the conversion factor of torr to atm, we can get that by...

Niharika Reddy 1D

Net change in moles of gas=Δn=(moles of gas in the products)-(moles of gas in the reactants)

For this problem, there are 3 moles of gas in the products and 2 moles of gas in the reactants, so the net change in moles of gas is 3-2=1 mol.

Niharika Reddy 1D

I think so, since K is included in the units of entropy. For enthalpy, when calculating q using q=mCΔT or a variation of that formula, the temperature could be left in Celsius if the heat capacity was also left in units of Celsius since relative temperatures made the change in temperature the same w...

Niharika Reddy 1D

I'm not entirely sure if this is how you do it, but this is how I approached the problem and got the same answer as the textbook. ΔH vap is the energy as heat required to vaporize a mole of benzene (units: kJ/mol). We are given the amount of energy required to vaporize 39.1 g of benzene (15.4 kJ). C...

Niharika Reddy 1D

If you look at the graph you plot in the beginning of the problem, it might make more sense. Work is not a state function, so it is path dependent. For path 1, there are 2 segments: one with a change in volume from 1.0 to 4.0 L at a constant pressure of 5.0 atm, and the second with a change in press...

Niharika Reddy 1D

Considering N₂ is diatomic and linear, I would go with the solutions manual errors document that uses the molar heat capacity at constant volume for a linear molecule: 5/2 R. N₂ is not monatomic so it cannot be 3/2 R.

Niharika Reddy 1D

Someone actually asked the same question a few questions down, but page 265 in the textbook says most thermochemical data are reported for 25° C, which is 298.15 K. It also says all reaction enthalpies used in the textbook are for 298.15 K unless otherwise indicated. So yes, that's what we assume.

Niharika Reddy 1D

2HCl(aq)+ Zn(s) ---> H 2 (g) + ZnCl 2 (aq) Since HCl is a strong acid, it dissociates and we can write it as 2H + (aq)+2Cl - (aq). ZnCl 2 is aqueous and soluble so we can also write that in its dissociated form: Zn 2+ (aq)+2Cl - (aq). Substitute the dissociated forms into the initial reaction to gi...

Niharika Reddy 1D

I think we assume the temperature is 298 K when it is not stated because page 265 in the textbook says most thermochemical data are reported for 25° C, which is 298.15 K. It also says all reaction enthalpies used in the textbook are for 298.15 K unless otherwise indicated.

Niharika Reddy 1D

Yes, as you said, you essentially just look at units to see which form of the gas constant you should use. The conversion factor between joules and (L atm) is 101.325 J=1 (L atm). If you take 0.08206 (L atm)/(K mol) and multiply it by 101.325 J/(L atm), you are left with another equivalent version o...

Niharika Reddy 1D

The data given in the problem is the standard enthalpy of combustion of each of the products and reactants of the final reaction. The standard enthalpy of combustion is the standard reaction enthalpy for the combustion of one mole of substance. This is why those reaction enthalpies are given next to...

Niharika Reddy 1D

I think the problem you're looking at is 105, not 101, but I'm not sure. Enthalpy of combustion is the change in enthalpy per mole of substance that is burned in a combustion reaction under standard conditions. This is essentially the same as the standard enthalpy of a (combustion) reaction where on...

Niharika Reddy 1D

Unless stated otherwise (as in problem 115 where it says the products of the combustion reaction are gaseous), the products of the combustion of an organic compound are carbon dioxide gas and liquid water, and any nitrogen present is released as nitrogen gas (N 2 ). If any other products are formed,...

Niharika Reddy 1D

The problem states: the gases in the four cylinders of an automobile engine expand from 0.22 L to 2.2 L during one ignition cycle. Assuming that the gear train maintains a steady pressure of 9.60 atm on the gases, how much work can the engine do in one cycle? To solve this problem, use the equation ...

Niharika Reddy 1D

Monatomic gases don't have any rotational energy because they have nothing around which to rotate, so yes, we only account for their translational energy since their rotational energy is zero.

Niharika Reddy 1D

You can convert the atmospheric pressure into atm by using the conversion factor 1 atm=760 torr. After calculating the change in volume, find the amount of work by using w=-PΔV and add this to the heat to get the change in internal energy.

Niharika Reddy 1D

It's supposed to be C=C, not CRC, but this was corrected in the solutions manual errors document on the chem 14B website. I have linked it below.

https://lavelle.chem.ucla.edu/wp-conten ... rs_5Ed.pdf

Niharika Reddy 1D

If you write our the combustion reaction for octane and balance it, 8 moles of CO 2 are produced. (The stoichiometric coefficients of octane, oxygen gas, carbon dioxide, and water are 1, 25/2, 8, and 9 respectively). You still use the enthalpy of combustion of octane, -5471 kJ, but divide it by 8 si...

Niharika Reddy 1D

You can split the problem into 2 parts, finding the change in enthalpy associated with 1. melting the ice and adding it to the change in enthalpy of 2. raising the temperature of the liquid water to its final temperature: 20°C. For melting the ice, multiply the standard enthalpy of fusion (6.01 kJ/m...

Niharika Reddy 1D

The change in internal energy, ΔU = q+w. For an isothermal change, or change under constant temperature, ΔU = 0. In a), going from solid to liquid requires energy, as the solid state is more stable than the liquid state. This would be endothermic, so heat is required, making q positive. Since ΔU = 0...

Niharika Reddy 1D

The Zn2+(aq) comes from the product ZnCl 2 (aq), which can be seen as Zn2+(aq) and 2Cl-(aq). The enthalpy of formation of the solid zinc is zero since zinc is in its most stable state. The enthalpy of formation of 2 moles of HCl(aq) cancels out with the enthalpy of formation of 2 moles of chloride i...

Niharika Reddy 1D

That value is the enthalpy of sublimation of carbon, and it is found in the problem itself. Enthalpy of sublimation is the molar enthalpy change when a solid converts directly into a gas.

Niharika Reddy 1D

Those gas constant values are equivalent, but given in different units. The conversion factor between joules and (L atm) is 101.325 J=1 (L atm). If you take 0.08206 (L atm)/(K mol) and multiply it by 101.325 J/(L atm), you are left with another equivalent version of the gas constant: 8.3145 J/(K mol...

Niharika Reddy 1D

The cylinder is the system and the cooling system makes up the surroundings. Since the surroundings are absorbing heat, the system (cylinder) must be giving it off, or releasing it. q is negative when heat is released.

Niharika Reddy 1D

In the problem, it is stated that we should assume the enthalpy of combustion of gasoline can be approximated by that of octane. This value is -5471 kJ/mol.

Niharika Reddy 1D

The energy required to heat the water can be found by using the equation q=mCΔT. The mass is 350.g, the specific heat of liquid water is 4.184 J/(g°C), and ΔT is 100.0°C-25.0°C=75.0°C. Multiplying these values will give the energy required to heat the water: 1.098x10^5 J. I have attached a picture o...

Niharika Reddy 1D

The 3 ways are: 1) Using Hess's Law : add the enthalpy changes at each step of a multi-step reaction. 2) Bond Enthalpies: find the sum of enthalpies of bonds broken (reactants, endothermic, energy needed, "positive" bond enthalpies) and bonds formed (products, exothermic, energy released, ...

Niharika Reddy 1D

When converting to Kelvin, convert the initial and final temperatures first, and then find the change in temperature from the two temperatures in Kelvin. I'm not sure if this is what you did, but if you find delta T in °C first and then convert that to Kelvin, it won't give you the correct answer.

Niharika Reddy 1D

The enthalpy of a reaction is the sum of the bond enthalpy changes. Bond enthalpies are all positive because by definition, a bond enthalpy is the heat required to break a bond, which is an endothermic process. For the reactants, bonds are broken and energy is needed for this to happen (endothermic,...

Niharika Reddy 1D

When calculated, work is not equal to zero; it's just a really small number compared to the energy gained as heat. Due to significant figures, the change in internal energy ends up being the same value as the energy added as heat, but the system is still doing some work. Therefore I would say the en...

Niharika Reddy 1D

If the K value is greater than 10^-5, you shouldn't approximate; leave the "-x" or "+x" in the equilibrium expression when solving for x. If you are unsure whether or not to approximate, you can make the approximation by assuming the change (x) is negligible, but then going back ...

Niharika Reddy 1D

Dilution changes the molarity because molarity is (moles of solute)/(liters of solution). When a species is diluted, the concentration decreases since there is more volume of solution but the amount of solute remains the same. To find the new molarity after it is diluted, use the equation: M i V i =...

Niharika Reddy 1D

It depends on whether the reaction is endothermic or exothermic. Increasing temperature always favors the endothermic direction of the reaction. In an endothermic reaction, energy is required and can be considered a reactant. Increasing temperature (and thus energy) drives the reaction forward, so t...

Niharika Reddy 1D

Bond strength is the dominating factor when determining the strength of binary acids such as HF and HCl; we look at the size of F and Cl to determine HCl is the stronger acid because Cl is larger than F and the H-Cl bond is longer and weaker, making it easier for water to remove a proton (H + ). For...

Niharika Reddy 1D

Generally, oxides of nonmetals form acidic oxides (SO 3 ), oxides of metals form basic oxides (BaO), and oxides of the elements in and close to the diagonal line of metalloids typically form amphoteric oxides (As 2 O 3 and Bi 2 O 3 ), which can function as both an acid and a base. Page 129 in the co...

Niharika Reddy 1D

The full ligand name is diethylenetriamine. The lone pairs on the 3 nitrogen atoms allow the nitrogens to serve as bond sites and coordinate to the transition metal cation, making dien a tridentate ligand.

Niharika Reddy 1D

I think you can do this problem like the other ones we have done as well, with a few variations to how it's shown in the solutions. If you convert just the initial amount of PCl 5 to concentration and say the change is -x for the reactant and +x for the products, you can set up the equilibrium expre...

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