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Hue_Vo_1D

First is used at 25 Celcius, a shortcut.

Both will yield you the same answer, but only at 298 F. At any other temp, use the second.

Hue_Vo_1D

In any reaction systems, the slow step will always determine the rate of reaction, thus the rate law. The purpose of using Pre-Equilirbrium Approach is to substitute other terms with intermediate, which is not supposed to be in the Rate Law. Suppose you have two-step reaction and the rate limiting s...

Hue_Vo_1D

In the Pre-Equilibrium Approach does the slow step always determine the rate law? In any reaction systems, the slow step will always determine the rate of reaction, thus the rate law. The purpose of using Pre-Equilirbrium Approach is to substitute other terms with intermediate, which is not suppose...

Hue_Vo_1D

4.36 in Organic Chem Book

Why is the standard Gibbs free energy of activation always a positive value for all reactions?

Hue_Vo_1D

There would be no y-intercept (i.e the line will not cross y-axis at lnA) because theoretically, it's impossible to reach the absolute zero temperature of 0K. Also, because 1/T is the x-axis, 1/0 would be undefined.

Hue_Vo_1D

In the case of adsorption, a relatively small quantity of one substance would get attached to the surface of the bulk substrate as a result of various types of interactions. In regards to materials that we have learned in this class, adsorption is the process in which reactants sit on surface of cat...

Hue_Vo_1D

Rate of zero-order rxn = k Note that what we're plotting in zero-order reaction is Concentration of R vs Time , which yields a straight line with slope of m, represented by k. The reaction is independent of initial condition , so [R]-reactant concentration of that specific rxn will decrease at a con...

Hue_Vo_1D

The format should be that: The anode always goes on the left and cathode on the right . Separate changes in phase by | and indicate the the salt bridge with || . Other important notes: Add "Pt(s)"-Platinum, which an inert conductor that must be in the cell with the half reaction that has o...

Hue_Vo_1D

If the problem has several calculation steps, should we round off at every step before moving to the next or leave it off until the end to round just the final answer?

Hue_Vo_1D

The unit Volt might make you confused. 1Volt=1Joule/Coulomb->similar to other intensive value such as density (g/mL). Standard reduction potential is a fixed value. Note: this means that if you multiply the half reaction by a constant, you don't multiply E° by the same constant since E° is an intens...

Hue_Vo_1D

Can someone please explain why on #5 the answer has 2NH3 on the product side? When I did the question and canceled everything, I divided 2NH3 by 2 so that it would be the same as in the target equation N2H4 + H2 = NH3. This gave me an enthalpy of -105.1kJ/mol, instead of -151 kJ/mol. Thanks I don't...

Hue_Vo_1D

-Write out the equation for the combustion of propane and make sure to balance the eqn. -From the given standard enthalpy of combustion (kJ/mol), figure out how many moles of propane is burned in order to release 2580. kJ of heat.(use stoichiometry). -Convert moles to mass using MM. Just let me kno...

Hue_Vo_1D

Can someone show me the setup for #6? Please and thank you! -Write out the equation for the combustion of propane and make sure to balance the eqn. -From the given standard enthalpy of combustion (kJ/mol), figure out how many moles of propane is burned in order to release 2580. kJ of heat.(use stoi...

Hue_Vo_1D

BrianaBarr2A wrote:For number 6, I'm a little stuck on which equation to use. Could someone help me with this?

Use the equation for the combustion of propane.
C3H8 + 5O2 => 3CO2 + 4H2O (balanced)

Hue_Vo_1D

Can someone show me the setup for #6? Please and thank you! -Write out the equation for the combustion of propane and make sure to balance the eqn. -From the given standard enthalpy of combustion (kJ/mol), figure out how many moles of propane is burned in order to release 2580. kJ of heat.(use stoi...

Hue_Vo_1D

In an isolated system, neither matter nor energy is exchanged with the surroundings->no change in internal energy->deltaU=0

Got it! Thank you.

Hue_Vo_1D

From what we are given, the reaction in the calorimeter releases 3.50 kJ of heat, which resulting in a temperature rise of 7.32 C in the calorimeter . The first step in solving this is -3.50kJ/7.32 C, which gives us .478 kJ/C; this is the heat capacity of this one specific calorimeter that is used i...

Hue_Vo_1D

Generally, the main concept of this question is that the heat capacity increases with molecular complexity

-As more atoms are present in the molecule, there are more possible bond vibrations that can absorb added energy-(NO2 has one more oxygen atom compared to NO)

Hue_Vo_1D

This is briefly explained in the book (reference page-269). Heat capacity, heat divided by change in temperature, itself is an extensive property, meaning that the larger the sample, the more heat is required to to raise its temperature by a given amount. It is, in a way, "uncertain" as th...

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Search found 19 matches

Re: The different Nernst equations

Re: pre equilibrium

Re: Catalyst

Free Energy of Activation

Re: lnA: does this y-intercept exist?

Re: adsorption?

Re: Zero Order Meaning

Re: Cell Diagram

Sig Figs: Rounding Off

Re: Intensive property

Re: Quiz 1 Preparation Answers

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Re: Quiz 1 Preparation Answers

Re: Quiz 1 Preparation Answers

Re: Quiz 1 Preparation Answers

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Re: Homework 8.25

Re: Homework 8.29: Need Clarification [ENDORSED]

Re: Heat Capacity