## Search found 36 matches

Fri Mar 16, 2018 11:39 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Calculate standard potential, Gibbs free
Replies: 2
Views: 201

### Re: Calculate standard potential, Gibbs free

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...
Fri Mar 16, 2018 11:25 am
Forum: Calculating Work of Expansion
Topic: Midterm #4A
Replies: 1
Views: 232

### Re: Midterm #4A

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
Fri Mar 16, 2018 11:22 am
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: question 14.97
Replies: 1
Views: 356

### Re: question 14.97

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 ...
Fri Mar 16, 2018 11:00 am
Forum: General Rate Laws
Topic: 15.99
Replies: 1
Views: 145

### Re: 15.99

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...
Sat Mar 10, 2018 9:23 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Why aren't homogeneous catalysts consumed by the reaction? [ENDORSED]
Replies: 1
Views: 182

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

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...
Sat Mar 10, 2018 9:01 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Stability using standard enthalpy of formation [ENDORSED]
Replies: 1
Views: 330

### Re: Stability using standard enthalpy of formation[ENDORSED]

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...
Sat Mar 10, 2018 8:46 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Test 2, Question #7 [ENDORSED]
Replies: 2
Views: 325

### Re: Test 2, Question #7[ENDORSED]

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....
Sun Mar 04, 2018 4:53 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Rate Constant in Terms of B, C, D, etc.
Replies: 1
Views: 175

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

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...
Sun Mar 04, 2018 4:04 pm
Forum: Second Order Reactions
Topic: Linearization of a Second Order Reaction
Replies: 3
Views: 755

### Re: Linearization of a Second Order Reaction

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.
Sun Mar 04, 2018 3:58 pm
Forum: General Rate Laws
Topic: Independent of the rate
Replies: 2
Views: 223

### Re: Independent of the rate

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.
Sun Feb 25, 2018 4:31 pm
Forum: General Rate Laws
Topic: Gas Product
Replies: 3
Views: 237

### Re: Gas Product

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.
Sun Feb 25, 2018 3:42 pm
Forum: General Rate Laws
Topic: Writing the Rate Law
Replies: 3
Views: 236

### Re: Writing the Rate Law

No i don't think we have to
Sun Feb 25, 2018 3:28 pm
Forum: First Order Reactions
Topic: Pseudo Order
Replies: 1
Views: 156

### Re: Pseudo Order

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...
Sun Feb 18, 2018 11:03 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Electrochemical Series Definition
Replies: 4
Views: 315

### Re: Electrochemical Series Definition

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.
Sun Feb 18, 2018 10:48 pm
Forum: Balancing Redox Reactions
Topic: 14.1
Replies: 5
Views: 319

### Re: 14.1

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...
Sun Feb 18, 2018 10:20 pm
Forum: Balancing Redox Reactions
Topic: Writing cell diagram and reaction equation
Replies: 3
Views: 253

### Re: Writing cell diagram and reaction equation

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...
Sun Feb 11, 2018 10:36 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: 8.93
Replies: 1
Views: 176

### Re: 8.93

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
Sun Feb 11, 2018 10:30 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Water effect 11.115
Replies: 2
Views: 386

### Re: Water effect 11.115

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.
Sun Feb 11, 2018 10:24 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Delta S total [ENDORSED]
Replies: 2
Views: 272

### Re: Delta S total[ENDORSED]

ΔS total=0 when the system is at equilibrium or the process is reversible.
Sun Feb 11, 2018 10:18 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Isothermic
Replies: 4
Views: 348

### Re: Isothermic

Isothermal process: ΔT = 0
Isochoric process or isometric process: ΔV = 0
Isobaric process: ΔP = 0
Sun Feb 11, 2018 4:38 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 8.53
Replies: 2
Views: 158

### Re: 8.53

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.
Sun Feb 11, 2018 3:48 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: ∆S = n R ln (V2/V1 )
Replies: 1
Views: 620

### Re: ∆S = n R ln (V2/V1 )

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)
Mon Feb 05, 2018 12:14 am
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: 9.43
Replies: 2
Views: 228

### Re: 9.43

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)
Mon Feb 05, 2018 12:09 am
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: 9.31
Replies: 2
Views: 212

### Re: 9.31

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...
Sun Feb 04, 2018 11:55 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 11.77
Replies: 1
Views: 80

### Re: 11.77

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.
Sun Feb 04, 2018 11:50 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Equations
Replies: 3
Views: 236

### Re: Equations

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...
Sun Feb 04, 2018 8:16 pm
Forum: Van't Hoff Equation
Topic: Constants
Replies: 4
Views: 439

### Re: Constants

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.
Sun Jan 28, 2018 9:31 pm
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: 9.19
Replies: 2
Views: 244

### Re: 9.19

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 ...
Sun Jan 28, 2018 9:14 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.13
Replies: 3
Views: 217

### Re: 9.13

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 ...
Sun Jan 28, 2018 9:05 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.13 [ENDORSED]
Replies: 6
Views: 561

### Re: 9.13[ENDORSED]

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 ...
Sat Jan 20, 2018 4:31 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Problem 8.99
Replies: 6
Views: 353

### Re: Problem 8.99

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...
Sat Jan 20, 2018 4:27 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Specific Heat Capacity
Replies: 3
Views: 239

### Re: Specific Heat Capacity

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.
Sat Jan 20, 2018 4:05 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Chapter 8 #99
Replies: 1
Views: 116

### Re: Chapter 8 #99

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...
Sun Jan 14, 2018 5:08 pm
Forum: Calculating Work of Expansion
Topic: application of PV=nRT
Replies: 3
Views: 220

### Re: application of PV=nRT

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.
Sun Jan 14, 2018 4:52 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Result of Calorimetry Calculations
Replies: 1
Views: 142

### Re: Result of Calorimetry Calculations

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.
Sun Jan 14, 2018 4:35 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: q and w
Replies: 5
Views: 342

### Re: q and w

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...