CHEMISTRY COMMUNITY

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The reaction profile displayed in the solution manual represents a more likely scenario than the one you described. The third step will not be endothermic (products having more potential energy) because doing so necessitates that an intermediate (reactant in step 3) be more thermodynamically stable ...

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Differential rate laws express the instantaneous rate of a reaction as a function of reactant concentrations. Integrated rate laws express the concentration of reactant remaining as a function of time. You'll have 2k as a constant if you solve for the rate of consumption/formation of a particular sp...

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A reaction with an overall order of 3/2 requires that its slow, rate-determining elementary step has fractional stoichiometric coefficients. The rate law for a 3/2 order reaction would have a rate constant with units (L/mol)^1/2 * s^-1.

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To find the unique rate of a reaction, you take the rate (of any product or reactant) and divide by the stoichiometric coefficient of that chosen species Usually you'll first be asked to find the rate law, in which case you'll often use the method of initial rates. Recognizing that this rate law gi...

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In the first step, the solutions manual finds the concentration of A when the concentration of B has increased to 0.19 mol/L. The next step where 0.055(molA)/L = 0.37[A]0 is just a mathematical formalism expressing [A] in terms of [A]0. As you can see from the subsequent calculation steps, this seco...

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The graph of 1/[A] v Time for a second-order reaction yields a line with slope=k and y-intercept=1/[A]i. Hence the graph of [A] v Time for a second-order reaction would not be linear. The graph would be a rational function with a vertical asymptote at t= -1/(k[A]i).

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Rate laws can be differential and take the form -d[A]/(a dt) or k[A]^n for the reaction aA --> P. Differential rate laws give the rate of the reaction with respect to the concentrations of reactants. Differential rate laws give insight into the reaction profile. For example, the rate law R = k [A]^2...

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Consider the equation Rate = k [A]^n [B]^m. For a table of experiments containing data on the initial [A], [B], and rate for each experiment, use the method of initial rates to determine n, m, and k. Determine m and n by setting up ratios of Rate = k [A]^n [B]^m for different experiments such that e...

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Use the method of initial rates to determine the reaction order with respect to each of the reactants. Then sum those orders to obtain the overall reaction order. Alternatively, one can determine the overall reaction order from the plots of [A] vs time, ln[A] vs time, and 1/[A] vs time (the integrat...

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If the reaction is first order then the graph would be linear with a slope of -K. The same goes for zero and second order reactions. Their graphs should be linear. If the reaction is first order, then the plot of ln[A] vs time is linear and the slope equals -k. Similarly, for a zero order reaction,...

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Is there a way of knowing if the metal will be oxidized? Do certain metals oxidize easier than others? Which metals are least likely to oxidize and dissolve into solution? Refer to the ordered electrochemical series. Metals with half-reactions that appear later in the table (more negative E^0) are ...

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The maximum cell potential occurs at the instant when the anode and cathode are first connected and the flow of electrons begins. Beyond that point, the concentrations in the half-reactions change. The production of product decreases the cell potential per increasing Q in the Nernst equation: E= E^0...

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I assume you are referring to the reactions posed in parts a and b of question 6L.7. Balancing these reactions requires some additional insight because the species involved in the redox half-reactions cancel in the full balanced equation. For the bronsted neutralization reaction, you need to add O2(...

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The balanced redox reaction should be: MnO4-(aq) + 5Fe^2+(aq) + 8 H+(aq) --> Mn^2+(aq) + 5Fe^3+(aq) + 4H2O(l). As for the galvanic cell, we know our cathode and our anode from the balanced half-reactions: a) MnO4-(aq) +8H+(aq)+5e- yields Mn2+(aq) +4H2O (cathode half reaction) 5[Fe2+ (aq) yields Fe3+...

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The question asks you to order the metals starting with the most strongly reducing metal. The "most strongly reducing metal" is the strongest reducing agent, which is the metal half-reaction with the most negative E^0 value (in this question, Al). You are right that the more positive the E...

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Absolute potentials cannot be measured, so scientists need a reference electrode. The selection of hydrogen for this purpose is arbitrary. Early researchers of electrochemistry likely selected hydrogen given the relative ease of obtaining/constructing a hydrogen electrode/solution.

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In S2O3 2-(aq), S has an oxidation number of +2. In SO4 2-(aq), S has an oxidation number of +6. After balancing the oxidation half-reaction, SO4 2-(aq) has a stoichiometric coefficient of 2. Since S2O3 2-(aq) has 2 S atoms, each with an oxidation number of +2, those S atoms together have an oxidati...

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Although oxygen's oxidation number doesn't change from O3 to O2, it changes from O3 to BrO3 2- (aq). O has an oxidation number of -2 in BrO3 2-, so oxygen gains 2 electrons (reduced).

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The salt used in a salt bridge should not form a precipitate with the ions in either of the half-reaction cells. Thus Cl- would actually not be a good salt when analyzing the redox potential for Ag+(aq) + e- --> Ag(s) because Cl- reacts with Ag+(aq). Provided K+ or Cl- does not form a precipitate wi...

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Start by identifying the oxidized and reduced species. The oxidation number for Cr changes from +6 to +3, indicating reduction (gain of 3 electrons). The oxidation number for C changes from -2 to -1, indicating oxidation (loss of 1 electron). You now have the skeleton for your half-reactions: (Cr2O7...

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The ΔS of the irreversible process is the same as that of the reversible process because entropy is a state property. The pathway is irrelevant.

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Even if you balanced the equation with whole-number coefficients, you are still only reacting 1.00 mol C6H6 (not 2.00 mol). Calculating the moles of gaseous reactants/products using 1.00 mol C6H6 will give the same delta n regardless of which balanced equation you use.

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Yes, you need to use delta S= n C ln(T2/T1). To calculate the standard entropy of vaporization of water at 85 C, you need to break the problem down into four steps: 1. Calculate the entropy change as water heats from 85 C to 100 C using delta S= n C ln(T2/T1). 2. Record the entropy of vaporization a...

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There are 15 ways to group two items from a set of 6 (6 choose 2). This is the total number of configurations of a pair of distinct atoms around an octahedral molecule, so #cis configurations + #trans configurations = 15. There are 3 trans configurations because there are 3 ways to group two atoms (...

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1 mole of atoms making up diatomic molecules equals 0.5 moles of diatomic molecules. 1.0 mole of a monoatomic gas has a greater number of microstates than 0.5 moles of diatomic molecules because the monoatomic gas involves a greater number of particles.

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Estimate the boiling point from (delta Svap) = (delta Hvap)/T where delta Svap= 85 J K^-1 and delta Hvap = 21.51 kJ/mol. The calculation yields a boiling point of 253 K. The actual boiling point is 248-249 K. Trouton's rule states that roughly the same increase in positional disorder occurs when any...

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CristinaMorales1F wrote:Also, Cp = 5/2R and Cv=3/2R are only used for monoatomic ideal gases.

For linear molecules, use Cp = 7/2 R and Cv= 5/2 R to account for rotational kinetic energy along two axes.
For nonlinear molecules, use Cp = 4R and Cv = 3R to account for rotational kinetic energy along a third axis.

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Heat is the transfer of energy resulting from a temperature difference between a system and its surroundings. At the molecular level, heat is the transfer of kinetic energy between particles via collisions.

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Emily Chirila 2E wrote:Degeneracy is the number of ways of achieving a given energy state. W=2^n where n= # of particles

W = 2^n for a two-state system. In general, W = m ^n where m is the # of possible states of a single molecule and n is the # of molecules

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For an octahedral molecule made of two atoms (A and B) attached to the central atom: 6 x A, 0 x B: 1 microstate because the molecule is symmetrical in all orientations 5 x A, 1 x B: 6 microstates because atom B can occupy each of the six positions attached to the central atom 4 x A, 2 x B: 15 micros...

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Yes, start by calculating the enthalpy change using the standard enthalpies of formation (use table 4D2 and the TNT formation enthalpy value provided in the problem). You should get -13168.48 kJ for the reaction. To calculate the number of liters of TNT that produces that enthalpy change, put on you...

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For a reversible expansion, all changes in volume occur in tiny steps. We calculate the work using the definite integral from v1 to v2 of - P dV. To derive an equation for constant pressure conditions, set P as a constant. The definite integral evaluates to - P delta V, the work equation for constan...

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Internal energy is a state property, so we need not consider the reaction's mechanism. This problem first asks you to calculate the calorimeter constant. Using qcal = Ccal delta T, you can plug in -3.50 kJ for qcal and 7.32 C for delta T. Solve and get Ccal = 0.478 kJ C^-1. Since we now know the hea...

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I'm confused about why the change in internal energy for the isothermal expansion of a gas is 0. I understand that the kinetic energy of the system isn't changing due to the constant temperature. But wouldn't the potential energy of the gas decrease because of reduced intermolecular interactions/rep...

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Both of these are used to find enthalpy/delta H but also find change in internal energy which is q + w. Therefore, q can either be q p or q v depending on the given conditions. Enthalpy/delta H is equal to qp, the amount of heat released or absorbed at a constant pressure. Using qv implies that the...

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The textbook stresses that "the greatest work is done for a process that takes place reversibly." I'm confused as to why this is always the case.

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An open system can exchange both matter and energy with the surroundings. A car engine is an open system because it exchanges heat and matter (carbon dioxide, carbon monoxide, water, and other byproducts) with its surroundings. An isolated system can exchange neither energy nor matter with its surro...

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The equation deltaH = deltaE + delta(PV) describes the change in enthalpy of a system. We need not measure the enthalpy at constant pressure per se but doing so simplifies the equation to deltaH = q (after some algebra). As a result, we know the heat (q) released or absorbed in the reaction equals t...

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Glucose is a great source of chemical potential energy. The oxidation of glucose releases 2880 kJ of energy/mol, 2823 kJ of which is available to do work. Although eukaryotic cells cannot harness all of this energy, cellular respiration is still ~34% efficient in converting the chemical energy of gl...

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We measure the internal energy change instead of the absolute internal energy because the absolute internal energy is unknowable, encompassing the energies of all the atoms (nuclei and electrons) in the system. As Maya stated, delta U is useful because it quantifies work and heat flow through the sy...

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The ideal gas law is a limiting law that performs well as P approaches 0. High pressures highlight its shortcomings. Van der Waals equation, another approximation of gas behavior, is more accurate because it accounts for the volume of the gas molecules and their intermolecular attractions. However, ...

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Hi Gabby. Yes, I2 is not included in the equilibrium expression K= (PCl2)(PHI)^2/(PHCl)^2. Since we know the K value, the initial partial pressure of HCl, and the stoichiometric coefficients, we can use an ICE table to solve for the equilibrium partial pressures. Set up an ice table with the initial...

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Hi Sara. Reducing the volume increases the concentrations of both X2 and X (concentration = n/V). According to the reaction quotient Q = [X]^2/[X2], products increase relative to reactants as a result of the concentration increases. Q > K and the system favors the reverse reaction, yielding more X2 ...

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Lavelle explained that solvents/liquids are excluded from equilibrium expressions for dilute solutions because their concentrations remain essentially unchanged throughout the reaction. When would a solution not be considered "dilute" and no longer treated as a pure substance? Is there a s...

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As Justin explained, compression affects the reaction by increasing the concentrations of gaseous reactants and products. The decrease in volume serves to multiply all concentrations by some factor. Since all concentrations increase by the same factor, the side with more moles experiences a greater ...

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The rates of the forward and reverse reactions are equal at equilibrium; they are not 0. The reactions, therefore, do not stop.

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What is the reasoning behind raising the product/reactant concentrations to their stoichiometric coefficients in the reaction quotient/equilibrium expression?

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The equilibrium expression (Kc) does not include solid reactants/products because solids do not have concentrations. The expression neglects liquids/solvents because they are usually present in such excess that their concentrations are essentially unchanged.

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If you are interested, Professor Lavelle shows the temperature-dependent change in equilibrium for the 2NO2 <--> N2O4 at the end of the Chemical Equilibrium Part 4 video module (starting at 47 minutes). As temperature increases, the solution appears dark amber/brown because of an increase in NO2 con...

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No, a higher pressure does not affect a reaction's equilibrium constant. The system does not respond to changes in pressure directly. Take the example of increasing the pressure by decreasing the volume, which changes the concentrations (n/V) of the reactants and products. It is this change in conce...

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