CHEMISTRY COMMUNITY

Sarkis Sislyan 1D

Also, the work for a reversible expansion is greater than the work for an irreversible expansion.

Sarkis Sislyan 1D

Idk why this is in 14b, but pH is the -log[H+], so if we take the 10^-pH, we get the [H+]

Sarkis Sislyan 1D

Remember heat does not equal temperature. Isothermal means temperature is held constant, but heat is not restricted. As a result, deltaU equals 0. When we plug this into the first law of thermodynamics equation of deltaU=q+w, we get q=-w.

Sarkis Sislyan 1D

In mechanisms, everything that comes after the slow-step is not used when determining the overall rate law of the proposed mechanism. It's only when there's a step before the slow, rate-determining step that we may use the pre-equilibrium approach.

Sarkis Sislyan 1D

The Pseudo-rate law is used whenever there is a reaction in which there are multiple reactants affecting the rate of the reaction. For example, if rate=k[A]^N[B]^M[C]^L, then we would need to use the pseudo-rate law in order to calculate the orders of the individual reactants. We do this by saying t...

Sarkis Sislyan 1D

How do we know when we should switch our units from mmol to mol? For example, in question 15.17, the solutions manual keeps the concentrations of A and B as well as the rate in terms of mmol, but in question 15.19, the concentrations of A, B, and C as well as the rate are converted into mol. Is ther...

Sarkis Sislyan 1D

In the textbook, there is an example of the Rate law of the decomposition of ozone, Rate=k [O3]^2[O2]^-1. It says that some species, typically products, do not have positive orders, meaning an increase in concentration correlates to a decrease in rate. Are we going to be tested on this concept?

Sarkis Sislyan 1D

Unless experimental data is given to you with which you can find the reaction order, the reaction order will be given to you.

Sarkis Sislyan 1D

The order of a reaction is found by adding up the orders of all the reactants. To find the order of a reactant, we can use experimental data given to us. If an increase in concentration of a reactant doesn't change the reaction rate, then that reactant is a 0 ordered reactant. If the ratio of concen...

Sarkis Sislyan 1D

The Nernst equation is used when finding the E under non-standard conditions. There are multiple versions of the equation, but they all mean the same thing and can be used interchangeably. I believe the equation you are talking about is E°=(RTlnK)/nF, which is pretty much the same thing again. This ...

Sarkis Sislyan 1D

The half-reactions given in the appendix of the textbook give the reduction reaction. Since the anode is where oxidation takes place, we would need to flip the half-reaction given in order to get the oxidation half-reaction. The cathode is where reduction takes place, so no such flipping is needed.

Sarkis Sislyan 1D

If Q>K then that means there is more product than at the equilibrium concentration. As such, LeChatelier's principle tells us that the reactants will be favored. If Q<K then that means there is more reactant than at the equilibrium concentration, so the products will be favored.

Sarkis Sislyan 1D

We use Au(s) for the anode because it needs a solid conducting metal for the electrode. This question is an example of a concentration cell, where both the anode and cathode are the same conducting metal.

Sarkis Sislyan 1D

The reducing agent is the one that gives away its electrons, causing the other molecule to be reduced, while the oxidizing agent is the one that receives electrons from the other molecule, causing the other molecule to be oxidized. As a result, the molecule getting oxidized is the reducing agent whi...

Sarkis Sislyan 1D

I'm not exactly sure what you're asking but in Cl-, the charge is -1. However, in Cl2 gas, the Cl's are neutral, 0 charge. As a result, we get the half reaction Cl2 + 2e- --> 2Cl-. We have 2 Cl- ions because Cl2 gas has 2 Cls and we need to balance both sides of the reaction

Sarkis Sislyan 1D

We would use H+ in these situations

Sarkis Sislyan 1D

The anode is the one that oxidizes and is usually on the left and cathode is the one that reduces and is usually on the right

Sarkis Sislyan 1D

Delta G=delta H - TdeltaS so if delta G=0 then either both delta H and delta S are zero or delta H equals T times delta S

Sarkis Sislyan 1D

S^o r is the standard reaction entropy

Sarkis Sislyan 1D

Enthalpy is heat under constant pressure, so wouldn't a change in pressure mean pressure isn't being held constant?

Sarkis Sislyan 1D

Also, the value of Cp as well as Cv will be given to us on exams.

Sarkis Sislyan 1D

Microstates are the positions molecules can take and are what we can use to determine entropy.

Sarkis Sislyan 1D

We get the equation w=- $nRTln\frac{V_{2}}{V_{1}}$ from the ideal gas law PV=nRT. We substitute P=(nRT)/V into the integral w= $-\int_{V_{1}}^{V_{2}}PdV$ and get w= $-\int_{V_{1}}^{V_{2}}\frac{nRT}{V}dV$ which gives us w=-nRT(lnV₂-lnV₁), which is rearranged as w= $- nRTln\frac{V_{2}}{V_{1}}$ .

Sarkis Sislyan 1D

Dr. Lavelle gave the example in lecture of a piston with internal pressure of 2 atm and an external pressure of 1 atm. With nothing being done to keep the piston in place (such as a pin being inserted), the internal pressure would push up against the weaker external pressure in a sudden movement. Th...

Sarkis Sislyan 1D

No, breaking bonds does not release energy, it requires energy to break. As a result, breaking a bond is an endothermic process.

Sarkis Sislyan 1D

Also remember $\Delta U=\Delta H+w$ when under constant pressure

Sarkis Sislyan 1D

Only diatomic molecules have accurate bond enthalpy measurements, all other molecules' bond enthalpies are taken as averages

Sarkis Sislyan 1D

The enthalpy of vaporization is given in the unit kJ/mol. In part (a), we are given the total heat required to vaporize the methane as well as the moles of methane present. In order to find the enthalpy of vaporization, all we do is divide the kJ's of heat required by the 0.579 mol methane to get th...

Sarkis Sislyan 1D

Bond enthalpy is not strictly the energy required to break the bond, it's also equivalent to the energy released when the same bond is formed. So remember that bond enthalpy is positive for the reactants (since breaking bonds requires energy) and that bond enthalpy is negative for the reactants (for...

Sarkis Sislyan 1D

Enthalpy of sublimation = H of vapor - H of solid

Sarkis Sislyan 1D

The liquid form of water has a greater heat capacity than both the solid and gaseous phases because in the liquid phase, the network of hydrogen bonds is allowed to be free, and these intermolecular bonds cause liquid water to need more energy(heat) to be broken.

CHEMISTRY COMMUNITY

Search found 31 matches

Re: Final Question W18: Reversible vs. Irreversible External Pressure

Re: pH

Re: Isothermal Changes

Re: Slow Step

Re: Pseudo-Reactions [ENDORSED]

Units in 15.17 vs. 15.19

Negative Orders

Re: Examples

Re: Ways of determining what order reactions are

Re: Nernst equation usage

Re: Writing Half-Reactions

Re: Q & K?

Re: Homework Question 14.13 Part (d)

Re: 14.5 c

Re: Cl2 and Cl -

Re: The use of H3O+ versus H+

Re: Anode and cathode

Re: delta G=0

Re: Entropy

Re: Free Energy and Pressure

Re: 9.7

Re: Micro states

Re: Homework 8.11

Re: Irreversible Expansion

Re: Bond enthalpies vs enthalpy of rxn - conceptual [ENDORSED]

Re: Delta U？

Re: Why is using bond enthalpies the least accurate method of finding reaction enthalpies?

Re: 8.37

Re: Is bond enthalpy the same as dissociation energy? [ENDORSED]

Re: Equations

Re: Specific heat of water