CHEMISTRY COMMUNITY

Kelly Seto 2J

A reaction with a very negative standard reduction potential means that that species wants to undergo oxidation, the reverse (hence negative) of reduction, and lose its electrons as much as possible. As it oxidizes, it then causes the other species involved to become reduced as it gains its electron...

Kelly Seto 2J

I'm not entirely sure what part of the problem this is in reference to, but -0.59 is the value you can get from
(38 kJ/mol / 8.314 x 10^-3kJ/mol K)(1/310 K - 1/298 K)
You don't need ln(A) values at all to solve this problem

Kelly Seto 2J

I believe Ka is the acid dissociation constant for HF

Kelly Seto 2J

A nucleophile is a chemical species that is attracted to positively charged particles; it's also in a sense the same as a Lewis base in that it acts as an "electron donor" to form bonds. I'm not sure about the second half of your question though

Kelly Seto 2J

Only the given experimental data should be taken into account when determining k and n with initial rates, and the coefficients should not play a role at all

Kelly Seto 2J

I would think it's always preferable to use ln since the first order reaction integrated rate law is based on the exponential decay equation that uses e

Kelly Seto 2J

The increase in the concentration of B is directly linked to a decrease in the concentration of A, and this decrease is modeled by (2 mol A/1 mol B)(.034 mol B/L), which is the amount of A that had to be used to produce the 0.34 M of B. Subtracting this amount from the initial concentration allows u...

Kelly Seto 2J

Always moles for concentrations

Kelly Seto 2J

In regards to the rate of formation of O 2 , first we find the rate of rxn for NO 2 which is the change in concentration over the change in time (130 mmol x L^-1 / 20 sec) giving us 6.5 mmol/sec The we see from the coefficients that the rate of formation of oxygen is 1/2 that of NO 2 , so it should ...

Kelly Seto 2J

Both molarity and partial pressures (in atm) can be entered into the reaction quotient, so for this it would be
(0.075M)² x 1 atm / (1.0 M)² x 1 atm, and any other value for partial pressure could be entered if stated in the problem as long as its in atm

Kelly Seto 2J

In this case, it means Tl⁺ is being both oxidized and reduced to become two different products, Tl³⁺ and Tl(s)

Kelly Seto 2J

E values are intensive properties, meaning that their values don't scale with the other components of the reaction

Kelly Seto 2J

I believe it's because many of the species are stated as being aqueous in these reactions, so the water is assumed to be present. Also water is neutral so it doesn't really contribute anything to the electron balance

Kelly Seto 2J

Yes because they contribute to the electron balance

Kelly Seto 2J

If the conditions are acidic you balance the half equations using H₃O⁺ ions and if they're basic you balance using OH^- ions

Kelly Seto 2J

You were correct in multiplying the second equation by 3, but you need to also multiply the third equation by three because the 6 e- produced in the 2nd equation are going to be part of the reactants in the 3rd equation, which in its current state only shows 2 e-. These two equations can be combined...

Kelly Seto 2J

C₆H₁₂O₆ + 6O₂ -> 6CO₂ + 6H₂O

Then Products minus Reactants

6(-393.51 kj/mol) + 6(-285.83 kj/mol) - (-1286 kj/mol)

Which gave me -2808 kj/mol

It seems that we did the same method, maybe one of your signs was just mixed up or something

Kelly Seto 2J

Can someone explain what Wmax means? Also why are delta G and Wmax equal at constant T and P? What is the relationship between the two?

Kelly Seto 2J

C is the just heat capacity of the system which we need to use in order to understand how much energy is needed to cause the temperature change in question. I don't think that there is a way to go about solving these without knowing the heat capacity for the system, however most of the questions wan...

Kelly Seto 2J

Let's look at our equation dG = dH - TdS We remember that dH = dU + PdV, so any increase in pressure will result in a larger dH value If a larger dH value is entered into the equation at constant temperature, we get dG = dH - dS resulting in a large dG, or free energy, value as a result Therefore, a...

Kelly Seto 2J

Extensive properties depend of the amount of substance or matter measured while intensive properties, like boiling point and density, do not

Kelly Seto 2J

The standard enthalpy of formation for this reaction basically means that 358.8 kJ are absorbed per mole of S 8 OR 4 moles of C OR 4 moles of CS 2 , so it can't be written as kJ/mol. However, since this reaction only uses one mole of S 8 , then it makes sense that it's (1.25 mol S 8 )(358.8 kJ/ 1 mo...

Kelly Seto 2J

I think Lavelle said that it was a typo in the book and that the O₂ in the second rxn is supposed to be 3/2, not 1, lol

Kelly Seto 2J

I think that it's just the nature of the textbook that when no explicit information about the conditions is given, it means that the reaction is occurring under standard conditions (298 K for thermodynamics)

Kelly Seto 2J

I'm sure there are multiple ways to do this, but what I did was: 2NO + O 2 -> 2NO 2 H = -114.1 kJ The standard enthalpy of formation for NO is 90.25 kJ and for O 2 , as a pure element, is 0 so 2(NO 2 ) - (2(90.25 kJ)+ 0 kJ) = -114.1 kJ // solving for NO 2 gives you 33.2 kJ Then: 4NO 2 + O 2 -> 2N 2 ...

Kelly Seto 2J

C_p is greater than C_v because at constant volume, no work is being done so C_p is just equivalent to dU, but under constant pressure, expansion work takes place so you have to add the energy used for work (PdV) to dU

Kelly Seto 2J

1.30 x 10^5 J is the amount of heat needed to raise the temperature of the water to 100.0 C 1.45 x 10^5 J is the total amount of heat needed to raise the temperature of both the water and the copper to 100.0 C, obtained by adding 1.30 x 10^5 J (water) with 1.48 x 10^4 J (copper). Then you can find t...

Kelly Seto 2J

No, I believe he will provide the necessary heat capacities and enthalpies on the tests

Kelly Seto 2J

I think if there is no temperature explicitly stated, 298 K is just the standard temperature used in thermodynamics

Kelly Seto 2J

Yes. Since work (at constant external pressure) is calculated by w= -P*delta V where delta V is (Vf-Vi): Any expansion will give us a positive delta V value as V final > V initial , resulting in a negative value for work. Similarly for compression, the delta V value will be negative as V final < V i...

CHEMISTRY COMMUNITY

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Delta G= Wmax

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