Search found 33 matches
- Sun Mar 18, 2018 10:36 am
- Forum: Balancing Redox Reactions
- Topic: test 2
- Replies: 5
- Views: 814
Re: test 2
you have to use delta G naught = -nFE for each step because you can't just add up the Es. when you get the final delta G, convert it back to E.
- Sun Mar 18, 2018 1:59 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Pre equilibrium approach
- Replies: 1
- Views: 434
Pre equilibrium approach
When solving for rate laws, would we have to use the pre equilibrium approach on the final? In section, I learned another way that made more sense to me so if we used that way, would we be penalized for it?
- Wed Mar 14, 2018 9:57 pm
- Forum: General Rate Laws
- Topic: Powers of Concentrations in Rate Law`
- Replies: 2
- Views: 417
Powers of Concentrations in Rate Law`
Since when writing rate laws, the coefficients of the equation are not the power in the rate law, like 3 I- is not [I-]^3, how would you determine the power of a concentration in a rate law? Is it impossible to determine just by looking at the chemical equation?
- Thu Mar 08, 2018 11:54 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.17
- Replies: 6
- Views: 1047
Re: 15.17
Since [C] is zero order, it is not listed in the rate law. rate=k[A][B]^2
- Thu Mar 08, 2018 11:52 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2989387
Re: Post All Chemistry Jokes Here
If a chemist is sick, and you can't helium, and you can't curium, then you might as well barium.
- Thu Mar 08, 2018 11:51 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2989387
Re: Post All Chemistry Jokes Here
Technically, alcohol is a solution.
- Thu Mar 08, 2018 11:49 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2989387
Re: Post All Chemistry Jokes Here
Today I told a chemistry joke, no reaction.
- Thu Mar 08, 2018 11:48 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2989387
Re: Post All Chemistry Jokes Here
Two chemists walk into a bar, First says "hello, I'll have H2O" Second says "I'll also have some water, and stop ordering like that, you sound like an idiot"
- Thu Mar 08, 2018 11:48 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2989387
Re: Post All Chemistry Jokes Here
I think I lost an electron. In fact, I'm positive.
- Thu Mar 08, 2018 11:47 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2989387
Re: Post All Chemistry Jokes Here
NaCl NaOH
RED ALERT! THE BASE IS UNDER A SALT!!
RED ALERT! THE BASE IS UNDER A SALT!!
- Thu Mar 08, 2018 11:46 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2989387
Re: Post All Chemistry Jokes Here
Did you know that you can cool yourself to -273.15˚C and still be 0 K?
- Thu Mar 08, 2018 11:41 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: HW 15.19a
- Replies: 3
- Views: 615
Re: HW 15.19a
Also I tried doing the first one over the third one and I also got 2 so it doesn't matter which one goes on top and which one is on the bottom. I just got decimals when I put the small one on top but the same answer - second order.
- Thu Mar 08, 2018 11:38 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: HW 15.19a
- Replies: 3
- Views: 615
Re: HW 15.19a
For [B], I used the first and third experiment since [B] is the only thing that changes then. So I write the third one over the first one: (\frac{3.02}{1.25})^{b} = \frac{50.8}{8.7} So, 3.02/1.25 = 2.416 Also, 50.8/8.7 = 5.839 I know that the order for [B] has to be a whole number so like 2 ...
- Thu Mar 08, 2018 11:27 pm
- Forum: First Order Reactions
- Topic: Pseudo-First-Order Reaction
- Replies: 1
- Views: 413
Re: Pseudo-First-Order Reaction
I believe the pseudo reactions are always going to be second order but one reactant's concentration is so high so when the one with the lesser concentration runs out, the big one barely changes so you can pretend that it didn't change and that its just a first order reaction with the reactant with t...
- Tue Mar 06, 2018 12:47 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 15.51
- Replies: 6
- Views: 908
Re: 15.51
So would I completely disregard the fast one?
- Tue Mar 06, 2018 1:14 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 15.51
- Replies: 6
- Views: 908
15.51
I was wondering why the rate law is the way that it is: Step 1: NO + Br2 --> NOBr2 (slow) Step 2: NOBr2 + NO --> NOBr + NOBr (fast) Write the rate law for the formation of NOBr implied by this mechanism. The answer in the back of the book is rate=k[NO][Br2] I understand that NOBr2 cancels and the ov...
- Mon Mar 05, 2018 12:18 am
- Forum: First Order Reactions
- Topic: k
- Replies: 16
- Views: 1563
Re: k
I'm sure that if they use a different unit in the problem like hours or minutes, you could leave the k in that unit instead of in seconds
- Mon Mar 05, 2018 12:15 am
- Forum: General Rate Laws
- Topic: zero order in rate laws
- Replies: 8
- Views: 1021
Re: zero order in rate laws
Since it ends up being 1, I wouldn't include it. It makes the equation look cluttered
- Mon Mar 05, 2018 12:14 am
- Forum: Second Order Reactions
- Topic: 15.35
- Replies: 2
- Views: 392
15.35
Not sure why I'm not getting the right answer The half-life for the second-order reaction of a substance A is 50.5 s when [A]0 = 0.84 mol*L^-1. Calculate the time needed for the concentration of A to decrease to (a) one-sixteenth....of its original value I used the the half life 2nd order reaction t...
- Thu Feb 22, 2018 1:10 am
- Forum: Balancing Redox Reactions
- Topic: 14.11
- Replies: 3
- Views: 672
Re: 14.11
I also have this question. In the solutions manual for the anode, they put O2(g) and H+(aq) on the same side of the equation and made H2O but I thought that they would be on opposite sides of each other in the chemical equation. Can anyone explain why this is?
- Mon Jan 29, 2018 11:11 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Entropies of Condensation
- Replies: 2
- Views: 301
Re: Entropies of Condensation
yeah it's just the sign is switched
- Mon Jan 29, 2018 11:09 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.27a
- Replies: 4
- Views: 453
Re: 9.27a
I think it's because HBr(g) is larger than HF(g) so less compact and would therefore have a higher entropy. This may be wrong so if anyone has a different explanation, please share it.
- Mon Jan 29, 2018 8:58 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: 9.19
- Replies: 4
- Views: 608
Re: 9.19
this problem is similar to example 9.6 on page 329, you have to add up the entropies from the three reversible steps of heating the water to 100 deg C, phase changing, then bringing it back down to 85 degrees C. All these added will be the same entropy of vaporizing water at 85 deg C in one irrevers...
- Sun Jan 28, 2018 7:56 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.13
- Replies: 7
- Views: 962
9.13
I was wondering about when to use Cp or Cv or whether to just use R. In 9.13, we have to assume ideal behavior: During the test of an internal combustion engine,3.00 L of nitrogen gas at 18.5 degrees Celsius was compressed suddenly and irreversibly to 0.500 L by driving in a piston. in the process, ...
- Wed Jan 24, 2018 1:34 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Closed vs. Isolated
- Replies: 8
- Views: 1064
Re: Closed vs. Isolated
A closed system can exchange energy with its surroundings but not matter. An isolated system can exchange neither matter nor energy with its surroundings.
- Wed Jan 24, 2018 1:28 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.73 (a) [ENDORSED]
- Replies: 2
- Views: 3129
8.73 (a) [ENDORSED]
I am having some trouble with 8.73(a). The question is "Use the bond enthalpies in Table 8.6 and 8.7 to estimate the reaction enthalpy for a) 3C2H2 --> C6H6" I looked up the molecular structure for both molecules. C2H2 has two carbons bonded with a triple bond and hydrogens attached to the...
- Tue Jan 23, 2018 1:56 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 8.49
- Replies: 2
- Views: 318
Re: 8.49
You would use the formula delta U = q + w. q =deltaH. w = delta n*R*T. We assume it is at room temperature so you use 298K for T. n is 1 bc 3 moles of gas are produced and 2 moles of gas were there initially. The enthalpy is in kJ so don't forget to convert it into Joules when adding to the w bc the...
- Sat Jan 20, 2018 4:13 pm
- Forum: Phase Changes & Related Calculations
- Topic: Why does steam cause severe burns?
- Replies: 4
- Views: 445
Re: Why does steam cause severe burns?
I also have a question about this. Why do both water and steam exist at 100 F? Doesn’t water evaporate at 100 F so wouldn’t it always turn into steam at 100F?
- Sat Jan 20, 2018 3:56 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Constant Volume vs Constant Pressure
- Replies: 6
- Views: 534
Re: Constant Volume vs Constant Pressure
When volume is constant, Delta V is 0 because delta V is final volume - initial volume. The formula for work, stated above, would indicate that work is then 0. If pressure is constant, work could still be done, if volume changes as well
- Sat Jan 20, 2018 3:46 pm
- Forum: Phase Changes & Related Calculations
- Topic: Heat Capacity
- Replies: 6
- Views: 744
Re: Heat Capacity
I would recommend sticking with Joules throughout the calculation of a problem and converting to kJ at the end because for specific heat, the values are in Joules and you might accidentally use kJ and mess up
- Sat Jan 13, 2018 11:48 am
- Forum: Phase Changes & Related Calculations
- Topic: Standard Reaction Enthalpy
- Replies: 8
- Views: 781
Re: Standard Reaction Enthalpy
For standard enthalpy of formation, I know it is the formation of 1 mole of a substance using its elements in their most stable form, so is it just the pure elements in their standard state, or can it ever be molecules of different elements combined in their standard state?
- Sat Jan 13, 2018 11:43 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: enthalpy
- Replies: 2
- Views: 244
Re: enthalpy
Enthalpy is the internal energy plus pressure and volume. Because enthalpy is total potential energy of a system, you have to take into account how external pressure and the volume affect the system.
- Sat Jan 13, 2018 11:37 am
- Forum: Phase Changes & Related Calculations
- Topic: Intensive vs. Extensive
- Replies: 4
- Views: 422
Re: Intensive vs. Extensive
Heat capacity is an extensive property because, as it is the heat required to raise the temperature of an object by 1 deg. C, it is dependent on the amount of the object--kJ/deg. C. To get it to be an intensive property, you would divide the heat capacity by the amount of substance present.This is c...