CHEMISTRY COMMUNITY

Shally Li 2C

Rate of formation of B is equal to 1/2 rate of decomp of A because A has a coefficient of 2.

Shally Li 2C

sonalivij wrote:For worksheet 8 #4d I am getting 144 M/s instead of 140 M/s. Am I doing something wrong?

She rounded to 140 because you need to have two sig figs.

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According to the outline, you need to understand how the collision model accounts for the temperature dependence of reactions. So in collision theory, molecules must collide at a minimum kinetic energy in order to break bonds and form new ones. The higher the temperature, the higher the kinetic ener...

Shally Li 2C

In experiment 1 and 3, the concentration of A also changes so you cannot determine if the change in initial rate is because of the change in [A] or [B]. In experiments 2 and 3, only [B] changes ([C] does not affect rate), so you know that any change in rate is due to the change in [B].

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A unique rate is the change in concentration of a reactant or product divided by the stoichiometric coefficient. It is the same for each reactant/product in the reaction but is negative for the reactant and positive for the product.

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Rate laws are determined experimentally, so they will be given. You cannot determine rate law from chemical equation unless it is an elementary reaction.

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The cathode is the half-rxn with a greater Eo. The greater the Eo, the greater its reducing power, so for 14.15a,
AgBr+e- ->Ag+Br- has a lower reducing power in comparison to Ag+ +e- ->Ag, so it is the anode.

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Eo is measured under standard conditions and it is a measure of the electron-pulling power of a single electrode.

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A reducing agent is a compound that acts as an electron donor (is oxidized) and therefore reduces another compound. The reducing power is the ability of a compound to reduce another compound.

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In the textbook it says that when the standard potential is positive, the reduction half-reaction has a greater electron-pulling power, so the tendency for the half-reaction to occur is stronger, which is what I think he meant by the reactants are favored.

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Oxidation potentials should just be the reverse sign of reduction potentials.

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Up to what sections of chapter 14 in the textbook (6th ed) should we know for Test 2?

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You can see the change in moles from the balanced chemical equation.

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Use the stoichiometric coefficients to find reaction enthalpy. So delta H = 2(-395.72)-2(-296.83). Then to find the enthalpy for 0.030 mol, divide your delta H by two and multiply by 0.030.

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It should either be given or you could be given enthalpy. The definition of enthalpy is heat at constant pressure, so if you're given enthalpy, you know pressure is constant.

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You have to convert L atm into joules. 1 L atm = 101.325 J

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Degeneracy is how many microstates there are. By the number of states for the base of the degeneracy equation, it means the number of ways the molecule can be arranged. For example, for carbon monoxide, it can either be arranged C-O or O-C, which is why the base is 2 and w=2^N

Shally Li 2C

The heat capacity at constant pressure (Cp) is greater than Cv because when heat is added at constant pressure, the gas is allowed to expand and some of the heat is used as work. At constant volume, the gas cannot expand so no work is done and all the heat goes into raising the temperature of the sy...

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A reversible process occurs when a system is at equilibrium (P internal = P external). It can be reversed by an infinitely small change in a variable. For example, if the pressure inside a cylinder is about equal to the pressure outside of it, the piston can move in either direction if there is a sm...

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You need to set q(ice)= -q(water)
q(ice) consists of both the heat required to melt the ice at 0 degrees as well as the heat required to raise the ice from 0 degrees to the final temperature.
So q(ice)=mCs(Tf-0.0) + n x enthalpy of fusion
q(water is just mCs(Tf-45)

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Heat capacity is the ratio of the heat supplied to the rise in temperature, so heat capacity=q/delta T

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After setting up an ICE table you should get the expression 1.80=x^2/(3-x). Simplifying gives you the quadratic equation x^2+1.80x-5.4=0. Solving for x you should get x=1.59. Since the equilibrium concentration for PCl5 is 3-x, [PCl5]=1.41, [PCl3]=[Cl2]=1.59

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Bar is the standard unit of pressure but 1 atm is equal to approximately 1.01 bar so when calculating equilibrium constant they are about the same.

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We used the synthesis reaction because that is what is given in table 11.2 and the problem tells you to refer to the table for the value of Kc. If you solved for the decomposition of HI you would get 1/Kc since it is the reverse reaction.

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You can use an ICE table when you are given initial concentrations and you need to find the equilibrium concentrations. The products will not always start at zero but the values will be given.

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Adding an inert gas has no effect on the equilibrium concentration so it does not change the constant.

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Catalysts do not change the equilibrium constant and has no effect on the equilibrium composition. It speeds up both the forward and reverse reactions to the same extent, so dynamic equilibrium is not affected.

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You can use the approximation where x is negligible when you have a small K value.

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According to the textbook, x can be disregarded when it is less than 5% of the initial value. You can do this when you have a small K value.

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If you increase the concentration of reactants, it will form more product, however this doesn't mean that the concentration of product will be greater than the concentration of reactants at equilibrium. Partial pressure is the pressure exerted by a gas in a mixture and is expressed in atm, bar, Torr...

CHEMISTRY COMMUNITY

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Re: Rate of formation, rate in terms of loss

Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)

Re: Collision Theory

Re: HW Number 17 (6th ed.)

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Test 2 Textbook Sections

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Re: Chemical Eq Part #3 Post-Module Assessment

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Re: 6th Edition Ch11 Question 1d