## Search found 11 matches

Mon Feb 29, 2016 8:59 pm
Forum: Administrative Questions and Class Announcements
Topic: Video Competition
Replies: 66
Views: 24775

### Re: Video Competition

Saloni Patel Lecture 1
Mitali Doshi Lecture 3
Sun Feb 28, 2016 11:10 am
Forum: *Alkanes
Replies: 2
Views: 394

Adding on to the previous response, an example of a radical would be a fluorine atom by itself. It is electron deficient, since it needs one more electron to be satisfied, and is thus an electrophile. If you were to draw an arrow towards fluorine, it would be one headed to complete fluorine's octet.
Sun Feb 21, 2016 10:59 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Homogeneous and Heterogeneous Catalysts
Replies: 3
Views: 637

### Re: Homogeneous and Heterogeneous Catalysts

The terms homogeneous and heterogenous catalysts don't have to do with if they appear in reactants or products. Rather, they have to do with the phases of the catalysts. A homogenous catalyst is in the same phase as the rest of the reactants in the reaction, for example, a catalyst that is a gas in ...
Wed Feb 17, 2016 2:27 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Initial concentration rate math question
Replies: 1
Views: 2038

### Re: Initial concentration rate math question

I am not sure if this correctly responds to your question, but A(t)=0.78, and in the fraction it is over A(0), which is just 1.00, so that is why in the next step it is only ln (0.78) = -kt. It is not necessary to multiply by 0.78 by A(0), because this just gives you the percentage relative to the i...
Wed Feb 17, 2016 2:04 pm
Forum: First Order Reactions
Topic: finding t 15.37
Replies: 2
Views: 462

### Re: finding t 15.37

What you are doing is essentially the same thing as the textbook. Their solution looks different because they use the log rule which is that log(x)^n = nlog(x). Let's begin by moving the negative sign from the denominator of the fraction (the one you proposed) to the top. In this case, the log rule ...
Mon Feb 15, 2016 1:46 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Rate
Replies: 2
Views: 372

### Re: Rate

I agree with the previous response. If a lot of activation energy is needed for a reaction to occur, it will often be slower because the reactants need that high amount of energy in order to react, and this could take more time. However, if only a little energy is needed, it is easier for the reacta...
Mon Feb 15, 2016 1:41 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Forward Reaction
Replies: 3
Views: 509

### Re: Forward Reaction

In addition, we assume that at initial time 0, there are no products present, so the only substances that can interact are the reactants. The rate at which they react is the overall reaction rate, until there is enough product to induce the reverse reaction.
Mon Feb 15, 2016 1:40 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Rates of Reactions
Replies: 2
Views: 394

### Re: Rates of Reactions

I would like to clarify what you noted in the question, that the reaction rates are decreasing. However, I think the rate is negative not because the rate is decreasing, but because the concentration of reactants is decreasing. I agree with the previous response, that making the rate of the products...
Sat Jan 23, 2016 7:15 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Ccal = -q/T or +q/T????
Replies: 1
Views: 457

### Re: Ccal = -q/T or +q/T????

I think it's confusing because the q could be pertaining to either the system or the calorimeter. When it is the q of the calorimeter, Ccal=q(cal)/T. and q(cal)=-q(system). So you must pay attention to what the q is referring to. Hope this helps!
Sun Jan 17, 2016 11:22 am
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Qsystem = -Qsurroundings help
Replies: 2
Views: 1287

### Re: Qsystem = -Qsurroundings help

Adding on to the previous response, the reason why they are equal and opposite is that there is no other energy transfers taking place besides between the system and its immediate surroundings. We assume that there is constant pressure and volume, so in a perfect system, with nothing else acting on ...
Sun Jan 10, 2016 10:09 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Hess's Law element formation
Replies: 1
Views: 225

### Re: Hess's Law element formation

I believe that most elements we will be given will be in standard state since they will usually be diatomic molecules, which we know are in standard state. Also, the course reader says "most stable form of C" for the example, so we should be given the information if it seems ambiguous.