CHEMISTRY COMMUNITY

Jiun Yue Chung 2I

k' (rate constant prime) is simply the 'reverse' of k (the rate constant). This means that k' is the rate constant for the reaction of products going back to the reactants.

Jiun Yue Chung 2I

The rate constant is deemed a constant for a specific reaction at a specific, constant temperature. This differs from reaction rate because the rate of reaction is the speed in which a reaction is occurring. A good way to think about it is that concentration increasing or decreasing may change the r...

Jiun Yue Chung 2I

I think you can identify it as irreversible because for that question, it says the external pressure is constant. For reversible changes, however, external pressure is ever so slightly changing.

Jiun Yue Chung 2I

Generally if the concentration of a reactant doubles along with the reation's initial rate, it should be first order reaction. However, always do the arithmetic just to be sure :)

Jiun Yue Chung 2I

I'm not sure what you meant by "here". However, the units for zero order reactions are (mol/L^-1 x (time unit)), for first order reactions is (time unit)^-1, and for second order reactions is (L x mol^-1 x (time unit)^-1).

Jiun Yue Chung 2I

That's a good model to write rate laws after. But remember that some rate laws may omit some reactants and of course, the exponents (in this case m,n, and l) will be different. :)

Jiun Yue Chung 2I

Also, there may be scenarios where your products are gases and they escape such that the only thing you can accurately rely on to calculate the rate is the reactants themselves.

Jiun Yue Chung 2I

I don't think we do, especially since rate laws simply deal with the concentrations of substances

Jiun Yue Chung 2I

We aren't just looking at initial (or at least I don't think so). There are two different equations we use to find the initial rate and then we use the Integrated Rate Laws to find rates after the start of the reaction :)

Jiun Yue Chung 2I

In order to balance half-reactions, you often want to balance the O atoms in each first, which requires adding H2O on the other side of the equation. For instance, the half-reaction MnO4- -----> MnO2 has 4 O atoms on the left and 2 O atoms on the right. To balance, you add 2 H2O molecules to the rig...

Jiun Yue Chung 2I

Since Platinum is an inert metallic compound, it doesn't affect the cell reaction. Moreover, Platinum is a good electric conductor, so I guess it's often used in this redox reactions.

Jiun Yue Chung 2I

You add H+ ions to redox reactions in acidic solution and OH- ions to redox reactions in basic solutions.

Jiun Yue Chung 2I

If you look at example 9.1, it tells you that you can use that same equation to calculate the change in entropy when a system is being heated (reversibly). So if the question says that at constant temperature, a system is gaining some amount of energy in joules, it should be safe to use this formula.

Jiun Yue Chung 2I

I think that if delta G is negative, the reaction is indeed spontaneous leading to the products being at a more stable energy state than the reactants.

Jiun Yue Chung 2I

I remember from going to the midterm review session that unless the question specifically says at constant pressure, to assume using constant volume heat capacity. I think inuitively, however, that Cv should be used in the first question since yes, the pressure is the one that is changing. The secon...

Jiun Yue Chung 2I

I remember from discussion last week and hearing about the fact that entropy is always increasing in the universe. How is this so?

Jiun Yue Chung 2I

To add, K, the equilibirum constant determines the relative concentrations (and partial pressures) of the products and reactants of the reaction at equilibrium while the Q, or reaction quotient says the same thing but when the reaction is not at equilibrium. A large K indicates that the concentratio...

Jiun Yue Chung 2I

I'm sure sig figs are important when doing the calculations in the textbook.

Jiun Yue Chung 2I

An adiabatic system is one where heat and matter does not flow in or outside of the system itself. A great example of an adiabatic system is a thermal container that is closed (kind of like a really good thermos). This means that if the internal energy of the system were to change in some way, then ...

Jiun Yue Chung 2I

The way I would approach this problem is to see it in steps. First, you have to calculate the heat (q) released when the ice is melted and then calculate heat from the water changing its temperature in response to the ice cube's presence. After identifying the information you're given, you know that...

Jiun Yue Chung 2I

I think that a good way to think about this concept is to remember the picture of the gas contained in a flask (blocked by a valve that prevents it from diffusing to the rest of the flask). When you open the valve, the gas diffuses out into the rest of the flask and therefore, entropy has increased ...

Jiun Yue Chung 2I

You could do final minus initial, but that would give you a negative change in moles. The question is asking how much H2 was burned, which already implies that the system is losing X amount of moles. The answer is positive because you wouldn't say -X moles of H2 was burned, but rather that X moles o...

Jiun Yue Chung 2I

Bond enthalpy calculation is reactant minus products because if you think about it, bond enthalpy is defined as the energy it takes to break 1 mol of a bond. Using this, positive values would be given to broken bonds (in the reactants) and negative values for bonds formed. Technically, enthalpy of r...

Jiun Yue Chung 2I

I couldn't find the bond enthalpy of a Br-Br bond, so I just followed the enthalpy given on the example. After identifying the bonds broken and formed in the reaction as shown, you basically calculate the regular enthalpy of reaction using bond enthalpies. I'm not sure if you had stopped there (whic...

Jiun Yue Chung 2I

I believe there is a difference between the two. Bond enthalpy refers to the amount of energy it takes to break one mole of a bond. On the other hand, Bond Dissociation Enthalpy is referring to the enthalpy change as two atoms that were previously bonded retain the electron they put into the bond in...

Jiun Yue Chung 2I

All the replies above are pretty accurate. I also think it's important not to get mixed up between chemical properties and physical ones. Just remember that intensive and extensive properties are just like sub-categories of physical properties too (that is because properties like color, mass, and vo...

Jiun Yue Chung 2I

I believe the reason why reversible processes can output the maximum amount of work is, if you recall, a reversible process must be done slowly and this causes the system to lose energy as heat much less than an irreversible process. Irreversible processes tend to be done more quickly and therefore ...

CHEMISTRY COMMUNITY

Search found 27 matches

Re: K and K'

Re: Rate constant

Re: Midterm 4a

Re: First order

Re: Units for Rate Laws 15.3

Re: general rate laws

Re: Initial rates

Re: Writing the Rate Law

Re: Instantaneous Rate

Re: "Rules"

Re: Platinum electrodes

Re: Add H+ or OH-

Re: Second Law

Re: Stability and deltaG

Re: Cp v. Cv!

Entropy question [ENDORSED]

Re: Constant K and reaction quotient

Re: Sig Figs

Re: Adiabatic system [ENDORSED]

Re: 8.41

Re: entropy and spontaneity

Re: 8.115

Re: How to calculate bond enthalpy

Re: Textbook Example 8.13

Re: Bond Enthalpy vs. Bond Dissociation Enthalpy

Re: Extensive/Intensive

Re: Reversible vs irreversible