CHEMISTRY COMMUNITY

Cole Woulbroun 1J

Will we only be expected to find the rate-determining step using the pre-equilibrium method, or will we be expected to use steady state as well? If the latter is the case, how can one distinguish whether to use the pre-equilibrium or steady state method in a given situation?

Cole Woulbroun 1J

While an intermediate is produced then consumed, a catalyst is consumed then regenerated.

Cole Woulbroun 1J

You will sometimes be given the slow step, but if not, you can use the equilibrium constants and equivalent ratios for each step in order to substitute values in to this step's rate law. When the rate law, which now includes these substituted values, matches with the overall rate law, then the rate-...

Cole Woulbroun 1J

Do we only use kJ/mol if we have been given the value or calculated a reaction with only one mole of a particular reactant?

Cole Woulbroun 1J

How do you know whether to use the units kJ/mol or J/mol in thermodynamic calculations?

Cole Woulbroun 1J

In the homework problem 4I.3, the chanfe in the entropy of the surroundings is asked for, however it does not specify whether the process was reversible or irreversible. Will this information be specified if asked for on the final?

Cole Woulbroun 1J

Many reactions are thought to occur in steps, as they make the formation of particular products from particular products more plausible. An intermediate is a molecule, atom, or ion that is produced in one of these stepwise reactions and consumed in another one of these stepwise reactions.

Cole Woulbroun 1J

Proper orientation also helps increase the reaction rate.

Cole Woulbroun 1J

I am not exactly sure about a non-enzymatic catalyst's mechanism of lowering activation energy, but the principles of the two types of catalysts performing the process is likely similar. In an enzyme, the various chemical groups on the amino acids that are exposed to the substrate form non-covalent ...

Cole Woulbroun 1J

Larger temperatures can supply more heat energy to the reactions, bringing their measured energy closer to the activation energy required for the reaction to proceed.

Cole Woulbroun 1J

In the Arrhenius equation, A does not represent concentrations. Instead, it is a reaction-based constant that describes the spatial requirements for the reaction to occur. If k is large, meaning that the reaction rate will be large, then A will also be large. A large A tells us that the spatial requ...

Cole Woulbroun 1J

Yes, as once you get past termolecular, the odds of that number of molecules colliding becomes extremely slim. Even termolecular reactions are rare and often occur at low rates.

Cole Woulbroun 1J

What determines whether a reaction is zero, first, or second order? I would assume that it would be molar ratios but I've seen exceptions to this so I was wondering what the explanation is.

Cole Woulbroun 1J

Rate of consumption is the rate at which the moles of reactant is decreasing and rate of product formation is the rate at which product is being formed. The rate of the reaction is represented as 1/a(-d[R]/dt), which is equal to k[R]^n.

Cole Woulbroun 1J

The rate constant k is different based on the order of reaction. Because Rate = k{R}^n, the units of the [R]^n portion of the equation will change based on the order. Despite this change in the mechanism of the reaction, the units for Rate must remain as M/s, as this is what the rate is. Therefore, ...

Cole Woulbroun 1J

A question in the book asks what range (in volts) does a voltmeter need to have to measure pH in the range of 1 to 14 at 25 °C if the voltage is zero when pH = 7? I'm having trouble understanding how acid/base equilibria represents a concentration cell and how I would use equations like Nernst in th...

Cole Woulbroun 1J

What is meant by the term electrolyte? I've seen it in a lot of electrolysis problems in the book but never saw the definition in the context of electrolysis.

Cole Woulbroun 1J

Can you post your steps? In a neuron (a nerve cell), the concentration of K+ ions inside the cell is about 20–30 times as great as that outside. What potential difference between the inside and the outside of the cell would you expect to measure if the difference is due only to the imbalance of pot...

Cole Woulbroun 1J

In the problem 6.61, the potential difference is asked for when the concentration inside a neuron is 20x-30x greater than outside a neuron. I tried to use the Nerst equation so the I could plug in the given Q but got around 0.03 V when the answer was 0.09 V. Anyone know what I'm doing wrong?

Cole Woulbroun 1J

For 6N.3, the solutions manual said that E(cell) = +0.067 V, but I got -0.067V due to the Nernst equation, as E^o(cell) was equal to zero, meaning that E(cell) = -(RT/nF)lnQ. Can someone explain where I am going wrong?

Cole Woulbroun 1J

The electrons move in order to reach their lowest point of free energy, meaning that if there is a release of free energy with the redox reaction, it will be spontaneous. This is why dG and E can be connected through Faraday's constant and the number of moles of electrons. The electrons move through...

Cole Woulbroun 1J

For the problem O3 (aq) + Br- (aq) -> O2 (g) + BrO3- (aq), you are supposed to split into the oxidation and reduction half reactions. The half reaction of the species with bromine is pretty clearly the oxidation half reaction, but I was curious about how the reduction reaction happens between O3 (aq...

Cole Woulbroun 1J

When standard cell potentials are used, they are calculated with the reduction potentials of the cathode an anode half reactions with the equation E(cell) = E(cathode) - E(anode), correct? Did anyone else see a different, possibly incorrect answer on the solutions manual for question 6M.1?

Cole Woulbroun 1J

When calculating a standard cell potential, do you subtract the potential on the right from the potential on the left or subtract the potential of the cathode from the potential of the anode?

Cole Woulbroun 1J

Before doing problem 5.55, I was under the impression that when you had a temperature that wasn't 298.15 K, you used the Vant Hoff Equation to determine the equilibrium constant since it accounts for the effect that a temperature change has on Kc. But in 5.55 the dG = -RTln(K) was used. What is the ...

Cole Woulbroun 1J

Yes, maximum work is a theoretical expression. In actuality, you will likely have both a change in work and heat over the course of the reaction, as dU = q + w. Expressing the change in Gibbs free energy as maximum work is just a simpler way to describe it.

Cole Woulbroun 1J

Yes, they are the same because entropy is a state function.

Cole Woulbroun 1J

In the pizza roll problem on the isobaric, isochoric, and isothermal process, the answer for the total change in entropy was stated to be zero. However, I calculated a different answer by adding the changes in entropies at each step. Does anyone know why the answer is zero?

Cole Woulbroun 1J

A question on the pizza rolls review packet asked to find the change in enthalpy for the reaction C6H12O6 + 6O2 = 6H2O + 6CO2 using bond enthalpies. Will we be expected to know the structures of common organic molecules like glucose for the midterm?

Cole Woulbroun 1J

You will be told whether a reaction is occurring at constant pressure or volume, and based on that you use the corresponding specific heat. This specific heat will be greater for constant pressure, as more heat energy is required to raise the temperature due to the fact that some of this heat energy...

Cole Woulbroun 1J

They should specify if the process is reversible or irreversible, as you do use different equations. For calculating work, you would use w = P(delta V) for an irreversible expansion and w = nRTln(V2/V1) for a reversible expansion.

Cole Woulbroun 1J

If you knew delta S of the surroundings and the temperature at which this process is occurring, you could calculate delta H with the equation delta S (surroundings) = - delta H/T

Cole Woulbroun 1J

In one of the homework problems, a question asks whether or not the decomposition of various into their elements or more basic molecules will occur. One of these molecules that could potentially decompose is HCN. What would this decomposition reaction look like?

Cole Woulbroun 1J

But wouldn't there had to have been a positive change in entropy for that process to occur spontaneously?

Cole Woulbroun 1J

In some of the homework problems I have done so far, the entropy change of the surroundings was asked for after the entropy change of the system was calculated. The answer was always the same magnitude of the entropy change of the system with an opposite sign. This would mean that the total entropy ...

Cole Woulbroun 1J

Work of expansion occurs when the gas is doing work on an object(s), such as a piston, and work of compression occurs when an object(s) is doing work on the gas. Work of expansion has a negative sign and work of compression has a positive sign.

Cole Woulbroun 1J

Yes. If you are only given one measurement of pressure and not asked to solve for another, you can assume that pressure is constant.

Cole Woulbroun 1J

Internal energy is the energy within a system. You can solve for the change of internal energy with the equation dU = dH + PdV. In this equation, where d stands for delta, dH represents the change in heat energy of the system while PdV represents the work either done on the system or done by the sys...

Cole Woulbroun 1J

I'm not sure whether or not we will be told what type of system the system in a problem is, but the important things to remember are that open systems allow for a transfer of matter and energy, closed systems allow for a transfer of energy, and isolated systems allow for a transfer of neither. Anoth...

Cole Woulbroun 1J

You solve this problem by setting up a heat transfer equation, q(ice) = -q(water). Because no phase change occurs with the water, q simply equals mC(dT). But for the ice, a phase change does occur. Because of this situation, q(ice) is composed of both the enthalpy of fusion and the mC(dT) equation. ...

Cole Woulbroun 1J

Does anyone know how to answer question 4A.9? I tried setting the q equations equal to each other to get the final tempertaure but my final temperature was lower than both the starting temperatures, which doesn't make sense. Thanks!

Cole Woulbroun 1J

This concept has to do with the intermolecular forces between water molecules. Because of hydrogen bonding between water molecules, water at 100 C has to have a significant amount of energy placed into it in order to overcome these interactions and vaporize into steam. This large energy requirement ...

Cole Woulbroun 1J

H can be viewed as a system's enthalpy, or its total internal energy. While this energy often takes the form of heat, this is not always the case. It can apply to other forms of energy as well. Since H represents energy, delta H represents the change in this energy over the course of a chemical reac...

Cole Woulbroun 1J

The two types of heat capacity are specific heat capacity and molar heat capacity. Both can can used to calculate q, however the variables in each of these equations change slightly to account for the different units each form of heat capacity uses. Whereas q=mCdT for specific heat capacity, q=nCdT ...

Cole Woulbroun 1J

Vaporization has such a large enthalpy because of the number of intermolecular interactions that must be broken in order for a gaseous state to be achieved. While melting only requires some intermolecular interactions to be broken, vaporization requires almost all intermolecular interactions be brok...

Cole Woulbroun 1J

A state property is a measurement that is not dependent on the path that was taken to get there. In a chemical reaction, there is a certain activation energy that must be achieved for the reaction to proceed. This activation energy is the highest point of energy for the entire reaction, but it is no...

Cole Woulbroun 1J

We use H3O+ because it describes the state of the dissociated H+ ions. After dissociating, H+ ions, or protons, create a covalent bond with one of oxygen's two lone pairs and, as a result, form a hydronium ion.

Cole Woulbroun 1J

It's possible to infer whether the reaction is exothermic or endothermic without knowing the exact delta H. Because bond formation releases energy, a reaction that forms bonds is exothermic. On the other hand, a reaction that forms bonds is endothermic, as this process requires energy.

Cole Woulbroun 1J

The ICE table can be calculated without using the quadratic formula when K is less than 10^-3. When K is less than 10^-3, one can reason that the change in concentration of the reactant, which has it's new concentration represented in the denominator, is insignificant and cross the x representing th...

Cole Woulbroun 1J

A gas is inert if it does not react. Only a few gases, such as neon, are inert, as these gases must have a full valence shell that prevents them from forming bonds. This lack of reactivity allows for inert gases to be placed in a container where a reaction in equilibrium is occurring and not effect ...

Cole Woulbroun 1J

When the pressure is changed through an increase or decrease in volume, the concentrations of the reactants and products change with it. This change in concentrations, either seen in a deficiency in products relative to reactants or reactants relative to products, causes the reaction to shift to a p...

Cole Woulbroun 1J

This problem requires you to set up an ICE box. Since NH4HS is solid, it is not included in the ICE box. However, it still reacts to produce H2S and more NH3 in addition to the 0.2 M of NH3 that was present before the reaction began. One you have added this unknown, additional amount of product to t...

Cole Woulbroun 1J

As long as K is set up correctly by taking into account the stoichiometric coefficients of the reactants and products, the value produced will be equal among these two reactions, as a reaction occurring at a single temperature has a consistent equilibrium constant.

Cole Woulbroun 1J

Are there multiple temperatures where a reaction can reach equilibrium or is there a set equilibrium temperature for a specific reaction?

Cole Woulbroun 1J

If one has a larger concentration of reactants than usual, the equilibrium concentration of products can be expected to be higher than usual as well, correct?

Cole Woulbroun 1J

If the concentration of a product or reactant is changed while a reaction is in equilibrium, will the reaction then be temporarily shifted out of equilibrium?

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