CHEMISTRY COMMUNITY

Alvin Tran 2E

The values are in Appendix 2A, but you have to scroll down to the organic compounds section. It starts on page A15.

Alvin Tran 2E

You can use both partial pressure and concentration in the equilibrium expression as long as the pressure is in bars and the concentration is in molarity.

Alvin Tran 2E

You'll be doing this when dealing with reaction mechanisms. Because an elementary reaction shows how the step of a reaction occurs, we can write the rate law of elementary reactions (not the overall reaction) with each exponent being the number of molecules of a particular type participating in that...

Alvin Tran 2E

emf is another way to mean the cell potential. They are the same thing.

Alvin Tran 2E

To find the order of [B] you want to compare reaction 1 and 3 because the concentrations of A and C stay the same. So we have \frac{Rate 3}{Rate 1} = \frac{50.8}{8.7} = \frac{k(1.25)^{N}(3.02)^{M}(1.25)^{L}}{k(1.25)^{N}(1.25)^{M}(1.25)^{L}} 5.84 = 2.41...

Alvin Tran 2E

The rate doesn't depend on the concentration of C, which means it is independent, because the reaction is zero order with respect to C. If you compare experiments 1 and 4 (where the concentrations of A and B stay the same), the initial concentration of C changes but the initial rate doesn't change.

Alvin Tran 2E

To find the order of [B] you want to compare reaction 1 and 3 because the concentrations of A and C stay the same. So we have \frac{Rate 3}{Rate 1} = \frac{50.8}{8.7} = \frac{k(1.25)^{N}(3.02)^{M}(1.25)^{L}}{k(1.25)^{N}(1.25)^{M}(1.25)^{L}} 5.84 = 2.41...

Alvin Tran 2E

If you compare experiments 1 and 4 (where the concentrations of A and B stay the same), you see that the initial concentration of C changes but the initial rate doesn't change. Therefore, the reaction is zero order with respect to C.

Alvin Tran 2E

That isn't the rate law but the unique rate of the reaction. In part A, you should have calculated the rate of reaction of NO 2 . In general for all reactions, aA -> bB + cC rate = \frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} The reactants and products in a re...

Alvin Tran 2E

I think it is because the half-lives of zero order and second order reactions depend on the concentration, so it is constantly changing as the concentration changes. In contrast, first order reaction half lives don't depend on concentration and as such are always constant, which makes it easier to u...

Alvin Tran 2E

To find the order of [B] you want to compare reaction 1 and 3 because the concentrations of A and C stay the same. So we have \frac{Rate 3}{Rate 1} = \frac{50.8}{8.7} = \frac{k(1.25)^{N}(3.02)^{M}(1.25)^{L}}{k(1.25)^{N}(1.25)^{M}(1.25)^{L}} 5.84 = 2.41...

Alvin Tran 2E

In addition, when you calculate the unique rate, it is the same for all the reactants and products involved in the reaction.

Alvin Tran 2E

R = 8.314 J / (mol x K) and T is in K.

Alvin Tran 2E

If the reaction depends on the concentrations of two or more reactants, you could say it is first order with respect to the reactant. So in the example in class today, it is first order with respect to NH 4 + (or in NH 4 + ) and first order in NO 2 - . And the overall order is the sum so it is a sec...

Alvin Tran 2E

The value of the rate constant K varies depending on the temperature and activation energy which can be seen through the Arrhenius equation. For example, raising the temperature of the reaction will increase the value of the rate constant K and therefore increase the speed at which the reaction is o...

Alvin Tran 2E

It doesn't matter. In the example in class, with rate 2 on top we got 2 = 2ⁿ so n = 1. If we do rate 1/rate 2, we get 1/2 = (1/2)ⁿ so n will still equal one.

Alvin Tran 2E

Adiabatic means there is no heat transfer so q = 0. From $\Delta U = q + w$ we get that $\Delta U = w$ .

Alvin Tran 2E

My TA said that posts are counted weekly, and you won't be able to make up points if you forget to post 3 times during a week.

Alvin Tran 2E

V = 11 L is given. n equals the sum of the He and Kr moles. It should be around 3.74 moles. R is the gas constant. T is the final temperature in K which is 75.0 + 273.15 = 348.15 K. Plug in and solve for P.

Alvin Tran 2E

You would use the specific heat of ice when it is in the ice phase. For example, say you want to find the q added to the system to get ice at -10 degrees celsius to water at +10 degrees celsius. You would find q needed to raise temperature from -10 to 0 degrees using specific heat of ice in the equa...

Alvin Tran 2E

Since the system is isothermal, the change in internal energy is 0. So q+w= 0, and q=-w. Since the balloon is expanding its volume against a changing pressure, work is being done. The energy lost during work is gained as heat, because the internal energy must equal 0.

Alvin Tran 2E

PV = nRT

Alvin Tran 2E

Heat is transferred because it is needed for a phase change from liquid to gas.

Alvin Tran 2E

On page 320 of the textbook it says, "A Note on Good Practice: Note the sign on the answer; always show the sign explicitly for the change in a quantity, even if it is positive." You wouldn't get marked wrong if you didn't have the positive sign though.

Alvin Tran 2E

It's good notation for whenever you're dealing with change in a value.

Alvin Tran 2E

C is oxidized from 2- to 1- the same way. C 2 H 5 OH has 0 charge. H has a charge of 1+, O has a charge of 2-. This gives H 5 OH a charge of 4. Each C in C 2 needs to have a charge of 2- to cancel out the 4+ charge. When you do this process for the product, C has a charge of 1- so it is oxidized fro...

Alvin Tran 2E

Cr₂O₇^2- has a charge of 2-. O usually has a charge of 2- and -2 x 7 = -14. So the charge of two Cr when added to -14 needs to equal -2. This makes the charge of Cr be 6+ because 2(6) -14 = -2.

Alvin Tran 2E

When you use the equation with specific heat capacity and constant volume or pressure you use (5/2)R. For monatomic ideal gases Cv=(3/2)R and Cp=(5/2)R. For diatomic gases, such as nitrogen in this problem, Cv=(5/2)R and Cp=(7/2)R

Alvin Tran 2E

You actually use C in the second equation too when dealing with temperature changes, but the different entropy equations come from ∆S=q/T so they basically function the same. C depends on whether you are calculating constant pressure or constant volume and is actually a multiple of R. In 9.13, you h...

Alvin Tran 2E

If you look on page 275 of the textbook, there is an example problem that shows that the pressure did drop by cooling. However, this takes place before any of the expansion work is actually done, so it is a vertical line. This line could have just as easily not been there, but it is just trying to s...

Alvin Tran 2E

The vertical brown line is the drop in the pressure. This was probably achieved by cooling at a constant volume (delta V = 0) so there is no work done.

Alvin Tran 2E

It should be a positive sign in the equation you gave, but you can get this equation from ΔU = q + w which is on the constants and equations sheet. Under constant pressure, ΔU = q_p + w. q_p = ΔH and w = -PΔV. If you plug those in and rearrange, you get the equation ΔH = ΔU + PΔV.

Alvin Tran 2E

What have you been getting for your answer? I got w = -0.94 kJ.

Alvin Tran 2E

Change in internal energy can be equaled to heat transfer when there is no work done (volume of reactants = volume of products). Change of internal energy is equal to work done when there is no heat exchanged. This can be seen in the equation ΔU = q + w.

Alvin Tran 2E

The equation ΔU = q + w comes from the fact that the internal energy of a closed system can be changed by heating/cooling or compression/expansion. Under constant pressure, ΔU = q_p + w. q_p = ΔH and w = -PΔV. If you plug those in and rearrange, you get the equation ΔH = ΔU + PΔV.

Alvin Tran 2E

Water and steam can both exist at 100 degrees F. However steam also contains more heat/energy than water because of the enthalpy of vaporization needed to turn water into steam. When steam comes into contact with the skin, it is exothermic and releases heat. This also releases the heat that was nece...

Alvin Tran 2E

The c makes it stand for enthalpy of combustion. You have to write a reaction for the combustion of each of the compounds (by adding O₂) so that you can find out their enthalpy of formation. From there, you can use Hess's Law to calculate enthalpy for the hydrogenation of ethyne to ethane reaction.

Alvin Tran 2E

Had to look up how a refrigerator works, but basically there is a compressor, and coil on the outside and inside of the fridge. Hot gaseous coolant is compressed and pushed into the coil on the outside of the fridge. When it meets the cooler temperature of a kitchen, it becomes a liquid. In liquid f...

Alvin Tran 2E

Also, the products side is negative because energy is released when new bonds are formed.

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