CHEMISTRY COMMUNITY

Anh Nguyen 2A

Because in 6d, its a galvanic cell so you use E=E cathode - E anode to calculate the overall standard potential of the whole cell. This has nothing to do with E not being state function. While in question 7, the reaction is made up of two steps (A turns into B and then B turns into C). For state fun...

Anh Nguyen 2A

You have to calculate Psys in order to determine that. For this question, Psys=1.1atm which is larger than Pext = 0.5atm so this is an irreversible expansion and w=-PΔV

Anh Nguyen 2A

F2(g) + 2H+(aq) + 2e- -> 2HF(aq) E1=3.03V F2 + 2e- -> 2F- E2=+2.87V (this one is from appendix 2B because you need a half reaction that has F-) Since Ka=[F-][H+]/[HF] the overall reaction has to be HF -> H+ + F- so when you calculate K, it's in the right order. And in order for that to happen, half ...

Anh Nguyen 2A

The general linear equation is f(x) = ax + b e. plot of k against T is not linear. If a reaction shows Arrhenius behavior then lnk against 1/T is a straight line, not k against T. f. rate=k[A] for first order reaction in A so the plot of rate against [A] is linear g. For zero order reaction in A, t1...

Anh Nguyen 2A

Homogeneous catalysts do participate in reaction. For example, 2H2O2 (aq) -> 2H2O(l) + O2 (g) this reaction has the catalyst Br2(aq) Step 1: Br2 is reduced to Br- Br2(aq) + H2O2(aq) -> 2Br-(aq) + 2H+(aq) + O2(g) Step 2: Br- is oxidized back to Br2 2Br-(aq) + H2O2(aq) + 2H+(aq) -> Br2(aq) + 2H2O(l) W...

Anh Nguyen 2A

The stability of a molecule depends on the spontaneity of its decomposition reaction. If its decomposition is spontaneous then that molecule is not stable. If the decomposition reaction is not spontaneous then that molecule is stable. The spontaneity of a reaction can only be assessed by ΔG naught a...

Anh Nguyen 2A

Cr2O72-(aq) +14H+(aq) +12e- -> 2Cr(s) +7H2O(l) This reaction is made up of 2 reactions: Cr2O72-(aq) +14H+(aq) +6e- -> 2Cr3+(aq) + 7H2O(l) E1=+1.33V 2Cr3+(aq) +6e- -> 2Cr(s) E2=-0.74V ΔG=ΔG1+ΔG2 <=> -nFE = -n1FE1 - n2FE2 <=> -12FE = -6FE1 -6FE2 <=> E = 1/2E1 + 1/2E2 <=> E= 1/2x(1.33) + 1/2x-0.74 = 0....

Anh Nguyen 2A

k is constant and characteristic of the reaction and the temperature so it does not matter if k is given in terms of loss of A or gain of C, etc., it would always be the same. I think that "given in the terms of loss of A" only clarifies what integrated rate law to use to find k. For examp...

Anh Nguyen 2A

The integrated rate law of second order reaction is 1/[A]t = 1/[A]o + kt which is in the form of a linear equation y= ax + b (b=1/[A]o and a=k) so graphing 1/[A] vs time would yield a straight line with the slope k and y-intercept=1/[A]o.

Anh Nguyen 2A

If the decomposition or the formation of that element is of a zero-order reaction then the concentration of that element is independent of the rate.

Anh Nguyen 2A

Because when the N2 leaves there would be less N2 in the system and the reaction would shift to the forward reaction to produce more N2 so reverse reaction is less likely.

Anh Nguyen 2A

No i don't think we have to

Anh Nguyen 2A

Some rate laws depend on the concentration of more than one species. In order to make the analysis easier, we can make the concentration of one species constant throughout the reaction. For example, one reaction has the rate r=k[A][B] and the concentration of B is constant throughout the reaction. T...

Anh Nguyen 2A

Electrochemical series is an arrangement of various metals in the order of their electrochemical activities based on the standard oxidation-reduction potential. The strongest oxidizing agents are at the top left of the table, the strongest reducing agents are at the bottom right of the table.

Anh Nguyen 2A

Cr 6+ in Cr2O7 2- is reduced to Cr3+ Reduction half reaction: Cr2O7 2-(aq) + 6e- + 14H+(aq) -> 2Cr3+(aq) + 7H2O(l) C 2- in C2H5OH is oxidized to C 1- in C2H4O Oxidation half reaction: C2H5OH(aq) -> C2H4O(aq) + 2e- + 2H+(aq) Since there are 6 e- in the reduction reaction and only 2e- in the oxidation...

Anh Nguyen 2A

Cathode is the electrode at which the reduction takes place. Anode is the electrode at which the oxidation takes place. There is no specific rule that assigns what is on the left or the right but it depends on the reaction. But usually we suppose that cathode is on the right and anode on the left. Z...

Anh Nguyen 2A

Since the work is done against the atmosphere, the pressure is constant or ΔP=0. In that case, w=-P.ΔV=-P.Δn(gas).R.T
with Δn(gas)=n(gas) products - n(gas) initial

Anh Nguyen 2A

Le Chatelier's principles only apply to gas. Since water is liquid, adding water to the system would have no effect on the reaction's equilibrium.

Anh Nguyen 2A

ΔS total=0 when the system is at equilibrium or the process is reversible.

Anh Nguyen 2A

Isothermal process: ΔT = 0
Isochoric process or isometric process: ΔV = 0
Isobaric process: ΔP = 0

Anh Nguyen 2A

Because the q you are calculating is q of the calorimeter so that's why there's no m. q cal=m.deltaT
q reaction = -q calorimeter, so if q calorimeter is positive then q reaction is negative.

Anh Nguyen 2A

For isothermal, reversible process (T is constant, delta U=q+w=0)
w=-n.R.T.ln(V2/V1)
q=-w => q=n.R.T.ln(V2/V1)
delta S=q/T
<=> delta S=(n.R.T.ln(V2/V1))/T
<=> delta S=n.R.ln(V2/V1)

Anh Nguyen 2A

You set qhot=-qcold to calculate the final temperature of the system as this is a thermally insulated vessel. (q=m.C.deltaT)
deltaS total = deltaS of the hot water + deltaS of the cold water
Calculate deltaS of each using deltaS=m.C.ln(T2/T1)

Anh Nguyen 2A

Actually you can know even when no information is given. Ethene (C2H4) is pretty small and light so you can safely say that it is in gas phase under normal condition while polyethylene is made up of many many ethylene molecules so it is heavy and solid. So ethylene has higher molar entropy than poly...

Anh Nguyen 2A

I don't think you need to know deltaS to do this. deltaH is larger than 0 which means this reaction is endothermic. So when the temperature increases, the reaction will shift towards the products.

Anh Nguyen 2A

Only at constant temperature and pressure, wmax=ΔG. The significance of this equation is that if ΔG of a process at constant temperature and pressure is known then we immediately know how much nonexpansion work it can do. Because in many cases, we are interested in many types of works, not just heat...

Anh Nguyen 2A

Because the values of enthalpy and entropy at different temperature are different but the difference between the final value and initial value is the same no matter what the temperature so we can assume that deltaH and deltaS are constant at different temperature.

Anh Nguyen 2A

I think you have to first know what changes are happening in this process. There are 3 steps to this process: 1. Reversible heating of water liquid from 85 degree C to 100 degree C For this you use the molar heat capacity at constant pressure of liquid water ΔS=n.Cp,m.ln(T2/T1) 2. Phase change from ...

Anh Nguyen 2A

I don't think we can combine those two processes in one equation because to calculate entropy change for change in temperature, we assume that the volume is constant (ΔS=n.Cv.ln(T2/T1)) and for change in volume, we assume that the temperature is constant (ΔS=n.R.ln(V2/V1)). They have to be separate ...

Anh Nguyen 2A

Because the process consists of 2 steps: 1.reversible isothermal compression at T initial from V initial to V final 2.Increase in temperature (T initial to T final) of the gas at constant volume (V final) That's why you use Cv when you calculate entropy change when temperature change. You can check ...

Anh Nguyen 2A

2HCl(aq) + Zn(s) -> Zn2+(aq) + 2Cl-(aq) + H2(g) The products of this reaction Zn2+, 2Cl-) and H2(g) . The enthalpy of formation of H2 is 0, Zn2+ is -153.89 and 2Cl- is 2(-167.16) => Enthalpy of formation of the product = (-153.89) + 2(-167.16) The reactants are 2HCl(aq) and Zn(s). The enthalpy of fo...

Anh Nguyen 2A

The K stands for Kelvin. Specific heat capacity is the energy needed to raise the temperature by 1 degree and since 1 degree of Celsius is the same as 1 degree of Kelvin, you can use it interchangeably.

Anh Nguyen 2A

2HCl(aq) + Zn(s) -> Zn2+(aq) + 2Cl-(aq) + H2(g) The products of this reaction Zn2+, 2Cl-) and H2(g) . The enthalpy of formation of H2 is 0, Zn2+ is -153.89 and 2Cl- is 2(-167.16) => Enthalpy of formation of the product = (-153.89) + 2(-167.16) The reactants are 2HCl(aq) and Zn(s). The enthalpy of fo...

Anh Nguyen 2A

You can calculate reaction enthalpies (ΔH) by using standard formulation enthalpies.
Change in internal energy is ΔU.
∆U=ΔH-Δn(gas)RT or you can replace Δn(gas)RT with PΔV.

Anh Nguyen 2A

The value you get from the calorimeter is q(cal) and q(cal)=-q.
So from the calorimeter, you can know the heat lost(or gained) by the reaction.
If you take q and divide it by ΔT, you would get the heat capacity (C).
Cm and Cs can be calculated by dividing C by moles and mass respectively.

Anh Nguyen 2A

q is the energy transferred by heat. w is the energy transferred by doing work. If heat has flowed from the system then q is negative. If the system absorbs heat then q is positive. And for w, if work is done on the system then w is positive, like compressing gas inside an insulated container increa...

CHEMISTRY COMMUNITY

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Re: Calculate standard potential, Gibbs free

Re: Midterm #4A

Re: question 14.97

Re: 15.99

Re: Why aren't homogeneous catalysts consumed by the reaction? [ENDORSED]

Re: Stability using standard enthalpy of formation [ENDORSED]

Re: Test 2, Question #7 [ENDORSED]

Re: Rate Constant in Terms of B, C, D, etc.

Re: Linearization of a Second Order Reaction

Re: Independent of the rate

Re: Gas Product

Re: Writing the Rate Law

Re: Pseudo Order

Re: Electrochemical Series Definition

Re: 14.1

Re: Writing cell diagram and reaction equation

Re: 8.93

Re: Water effect 11.115

Re: Delta S total [ENDORSED]

Re: Isothermic

Re: 8.53

Re: ∆S = n R ln (V2/V1 )

Re: 9.43

Re: 9.31

Re: 11.77

Re: Equations

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Re: 9.19

Re: 9.13

Re: 9.13 [ENDORSED]

Re: Problem 8.99

Re: Specific Heat Capacity

Re: Chapter 8 #99

Re: application of PV=nRT

Re: Result of Calorimetry Calculations

Re: q and w