CHEMISTRY COMMUNITY

Hubert Tang-1H

delta U is the change between the initial and final values of U, meaning that the pathway it took doesn't matter, as long as any path begins and ends on the same values.

Hubert Tang-1H

We first balance the equation into 2AB + C2 ---> 2ABC

Because C2 is consumed at a rate of 0.250 M/s, we can use the ratios to see that ABC is formaed at 0.500 M/s

We can then use
24.00 M-0.35 M = 23.65 M increase, and

23.65/(0.500 M/s)=47.3 seconds

Hubert Tang-1H

We split the reaction into the anode and cathode reactions: Anode: Au(s)--->Au3+ (aq) + 3e- Cathode: Au+(aq)+e- ----> Au(s) We see that both the anode and cathode have solid Au, so there shouldn't be Pt in the cell diagram. Thus, from our anode and cathode, our cell diagram should be Au(s)| Au3+(aq)...

Hubert Tang-1H

Because the ratio of SO2 to O2 is 2 to 1, when we use all 0.030 moles of SO2 for the reaction, we also use 0.015 moles of O2 as well. Thus,
there is still 0.015 moles of O2 left over, meaning SO2 is the limiting reactant.

Hubert Tang-1H

For part b, we aren't actually comparing the temperatures 298 K and 350K. We are actually simply doing part a again with T=350 K. Therefore,
we should have T=350 for both k1 and k2, and have

k1/k2 = Ae^(-75kJ/mol/RT)/Ae^(-125kJ/mol/RT)
Substituting R=8.314 and T=350 K, we get

e^17.1824 = 3x10^7

Hubert Tang-1H

I think that normally, the question will have the data be in such a way that k will generally be the same throughout several experiments with different concentrations. I'm guessing that if there is a noticeable difference in k values from each experiment, we might average them out.

Hubert Tang-1H

Since this is a first order reaction, we can use the first order half life equation

t_(1/2)=ln 2/k

We know that the half-life is 1000s, so what we need to do is calculate for k

k=ln 2 /t_(1/2)=0.693/1000 s = 6.93x10^-4 s^-1

Hubert Tang-1H

We also know that because the reaction is a first order reaction, we can utilize the equation:
ln [A] = ln [A_0] - kt, and plug in the equations for A_0 = .153 mol/L and A=.153-2*.034=.085 mol/L, and t=115 s

Hubert Tang-1H

The reason is because the reaction is a second-order reaction, meaning we must use the equation 1/[A]=kt+1/[A_0]. We notice that that for 16*[A]=[A_0], we have 16/[A_0]=kt+1/[A_0] kt=15/[A_0]. Compared to the half -life 2[A]=[A_0] 2/[A_0]=kt+1/[A_0] kt=1/[A_0]. Thus, we see that for second-order rea...

Hubert Tang-1H

Depending on the order of the reaction, we see that the units for k will be M/s divided by M^n, where n is the sum of the orders of the compounds the the reaction.
Take for example Rate=k*[A]^2 * [B]. Combined, the total order is 2+1=3, so we have the unit of k be M/s * 1/M^3, or 1/(M^2*s)

Hubert Tang-1H

Elementary reactions are essentially reactions where the reactants react to form the products directly, rather than forming any intermediate products.

Hubert Tang-1H

I think that, if it's not specified, we would orient the reaction so that the potential is positive.

Hubert Tang-1H

Since we can derive one equation from the other, there doesn't seem to be any impact if we use one or the other. It just depends on which one you feel more comfortable using.

Hubert Tang-1H

This problem is actually electrolysis, which is different from a Galvanic cell. In electrolysis at a cathode, the more positive potential is preferred, which is why Ni is chosen istead of So4

Hubert Tang-1H

i think that in the case of Hydrogen, we alter the other electrode half-reaction i such a way that its Standard electric potential becomes positive. In this case, the original cell has Zinc as the cathode, but because it has a negative E value -0.76, we reverse the reaction to have a positive E valu...

Hubert Tang-1H

Increasing surface area generally won't increase E and E°, since they depend on the substances making up the electrodes, not how much there is.

Hubert Tang-1H

Pt means platinum, and we include it in cell diagrams when there aren't any solid metals on either side. It's also not included in the reaction equation because it is inert.

Hubert Tang-1H

When we're balancing the solution for acidic, we use H+ and H2O, whereas in a basic solution, we balance the sides with OH0 and H2O.

Hubert Tang-1H

Because delta_U=0 due to the initial and final states being the same, we have w=-q, so essentially, we just need to find the energy of the work, which we can then use to find heat. We can do this by splitting the problem into 3 parts: 1. Calculate work done compressing 50.0 L to 20.0 L 2. Calculate ...

Hubert Tang-1H

Yes. The formula for work is W=-P*Delta_V, so if V decreases (compresses), Work is done on the system, so Work is positive. Likewise, if V increases, Work is negative.

Hubert Tang-1H

The main difference between the two laws is what they have to say about entropy. The second law states how a system favors going from lower to higher entropy. The third law states how temperature affects entropy by having entropy approach 0 as temperature approaches 0 (for perfect crystals)

Hubert Tang-1H

Yes, ∆G=-RTln(k) is the standard Gibbs free energy of reaction, whereas the ∆G=0 as equilibrium is the Gibbs free energy of reaction. The difference is that Standard free energy is the calculated by finding the difference of products and reactants in their standard states, whereas the ∆G=0 equilibri...

Hubert Tang-1H

The double bond for COF2 actually restricts the configuration for the molecule. However, COF2 still ultimately has more configurations than BF3 due to the presence of oxygen and 2 fluorine molecules around the central carbon atom, meaning COF2 has higher molar entropy.

Hubert Tang-1H

The calculate entropy, we need to calculate the entropy of raising liquid water from 85 C to 100 C, turning into vapor, and then cooling the vapor back to 85 C. The molar heat capacities will help in calculating entropy in temperature change, and you add in the entropy of vaporization to get the tot...

Hubert Tang-1H

0.38 J/g C is the specific heat of copper, meaning the total heat supplied includes heating both the water and the copper kettle.

Hubert Tang-1H

Kb is a constant, at 1.381 x 10^-23 J/K. When calculating W for 1.00 mole where there are n states for each molecule, you raise n to the power of 6.02x10^23 (Avagadro's Number), so for 1.00 mole, you can
have Entropy S= (6.02x10^23) x Kb ln n, where there are n states for the molecule.

Hubert Tang-1H

The equation delta_S=delta_q/T means there is a change in entropy due to a change in heat, rather than the overall entropy. Therefore, what this equation really means is that, at higher temperatures, a change in heat correlates to less of a change in entropy. Adding more heat does means more entropy...

Hubert Tang-1H

Cp means the heat capacity of a gas at constant pressure. We use this equation as part of the formula for calculating entropy in the problem, along with C_v, which is heat capacity for constant volume.

Hubert Tang-1H

I think the ideal conditions part means that the nitrogen gas is behaving as an ideal gas.

I don't know about the 1.00 moles of gas, though, but I don't think it has to do with the ideal conditions part.

Hubert Tang-1H

I think we can use the equation PV=nRT. Given a constant mole of Argon and Temperature, we see that Pressure and volume are inversely proportional, so because 2.00 atm is higher than 1.00 atm, the 2.00 atm situation will have a smaller volume.

Hubert Tang-1H

It is because the ice melts at 0.0 C in the problem, and becomes water. We therefore use the specific heat of water rather than ice.

Hubert Tang-1H

I think it may be because the volume has doubled, and in V1, the positions available consist of the entire volume of the container. V2 is double the volume of V1, meaning there is twice the volume available for gas molecules. I also think we assume that the gas is an ideal gas, so the volume occupie...

Hubert Tang-1H

R is the gas constant, which is 8.314 J/(mol*K)

Hubert Tang-1H

I would also like to know as well, though right now, I think we may be given a table on the information if a problem requires us to use this knowledge.

Hubert Tang-1H

We are measuring the internal energy of the system. Compressing the gas with 340 kJ of work. Remember that the Work done in a a system is w= - P*delta_V. The sign is negative, and because the system is compressed, delta_V is negative, meaning work done is positive, since compressing the system does ...

Hubert Tang-1H

I think that phase changes are only positive if the process is endothermic (melting and vaporization). The reason they are all positive is that the problem asks to convert ice into gas, which requires that the ice absorb heat, making the enthalpy positive.

Hubert Tang-1H

The heat needed to melt the ice cube and raise the melted cube's temperature comes from the surrounding water. Therefore, in actuality, we should calculate the heat needed to melt the ice cube and m*c*deltaT for ice, and equate it with the heat lost by the water in order to melt the ice cube. The eq...

Hubert Tang-1H

It's because we are calculating in terms of S8 moles rather than Carbon (We're calculating using the number of S8 moles rather than C). That means for 1 mole of S8 reacting in the equation, enthalpy change is 358.8 kJ. Our units is thus +358.8 kJ/1 mole S8, which we use to calculate the answer, rath...

Hubert Tang-1H

It's because both Celsius and Kelvin have the same "scale" (An increase of 1 Celsius is equal to 1 Kelvin).By finding the difference between the two temperatures, the conversion cancels out. If we convert 20 C and 100 C to Kelvin, we get 293.15 K and 373.15 K respectively, and finding the ...

Hubert Tang-1H

I've also gotten the answer of 448.5 kJ, rounded to 449 in my calculations. It doesn't seem to be due to be a rounding error, as you get the answer 448.5 kJ from only one equation. In addition, this situation also appears to occur in other answers I've gotten here and there. How accurate do we need ...

Hubert Tang-1H

Heat capacity and specific heat capacity differ in the fact that heat capacity only accounts for temperature and energy, whereas specific heat capacity also takes into account mass. As such, the units for heat capacity is Joules/Kelvin (or Celsius), whereas the units for specific heat capacity is of...

Hubert Tang-1H

Endothermic reactions absorb heat, and in most cases, this heat is taken from the surroundings. In this case, the flask is considered the surroundings of the reaction (that's why the exothermic reaction, which releases heat, warms the flask/surroundings). The endothermic reaction absorbs the heat fr...

Hubert Tang-1H

The most important phase changes would probably be between solids and liquids (melting and freezing) along with phase changes between liquids and gases (evaporation and condensation). I think sublimation (solid to gas) and deposition (gas to solid) i might also be talked about in class, though I'm n...

Hubert Tang-1H

For superheating, boiling typically occurs when the vapor pressure of a liquid equals the pressure of its surroundings. Some liquids may require additional pressure to overcome surface tension. Superheating basically occurs because of the surface tension and surrounding pressure both opposing the va...

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