CHEMISTRY COMMUNITY

Harrison Wang 1H

Catalysts are involved in the first step of the equation on the reactants side and then reformed in another elementary step as products. Intermediates are formed during an elementary step first as products and then re-used as reactants later in another elementary step. Both do not show up in the ove...

Harrison Wang 1H

The functional groups we learned today are part of the 14B syllabus and so I would expect that they are fair to be tested upon for the final

Harrison Wang 1H

Heterogeneous catalysts are usually only available in the solid state and therefore mainly function through adsorption, homogeneous catalysts are available in all states and therefore can but no do not have to function through adsorption.

Harrison Wang 1H

You can show your work by writing the ratio of concentration in exp 1 over concentration in exp 2 raised to some power equals the ratio of rate in exp 1 over rate in exp 2. This would be written as (conc1/conc2)^n=(rate1/rate2) where n is the order.

Harrison Wang 1H

For second order integrated rate law = k is positive, reciprocal of concentration is taken
For first order integrated rate law = ln/e will be involved, k is negative
For 0 order integrated rate law = k is negative, concentration is left as is

Harrison Wang 1H

Rate is given in units of mol/(Ls), and as such when doing calculations involving the rate law, all your values should be in moles.

Harrison Wang 1H

You do, as the orders of all reactants factor into the overall order of the reaction

Harrison Wang 1H

Zero - rate of reaction is equal to rate constant k
1st order - depends on only ONE reactant concentration
2nd order - depends on two different reactants or one reactant squared

Harrison Wang 1H

When determining the molar entropy of molecules, the phase of the molecule takes priority over size. That is, only if both molecules are in the same phase does size matter. Otherwise, the entropy of gases >> liquids > solids

Harrison Wang 1H

1. Identify the key element that undergoes an oxidation state change. 2. Balance the number of atoms of the key element on both sides. 3. Add the appropriate number of electrons to compensate for the change of oxidation state. 4. Add H+ (in acid medium), or OH- (in basic medium), to balance the char...

Harrison Wang 1H

For either the cathode or anode, if there is no conducting solid, then Pt(s) will be placed there. For example, if the reaction at the anode consists of Fe+2(aq) -> Fe+3 + e-, then Pt(s) will be placed on that side.

Harrison Wang 1H

A Galvanic cell converts chemical energy into electrical energy. An electrolytic cell converts electrical energy into chemical energy. Electrolysis is used to drive an oxidation-reduction reaction in a direction in which it does not occur spontaneously.

Harrison Wang 1H

The other molecules whose oxidation states remain the same throughout the reaction are spectators and don't have to be written in the final reaction.

Harrison Wang 1H

Standard change in free energy of a system at equilibrium is not 0, but rather related to the equilibrium constant K at that temperature. If K is >1, then standard change in free energy will be -, and if K<1, then standard change in free energy is +. It is delta g or change in energy that is 0 for a...

Harrison Wang 1H

I'm assuming that Test 2 will be on electrochemistry, which according to the electrochemistry outline spans from chapter 14.1 to 14.11.

Harrison Wang 1H

I think we need to understand that we get the standard reduction potential of elements from galvanic cells that involve the element in question and hydrogen. A cell that involves a standard hydrogen electrode is a considered a galvanic cell.

Harrison Wang 1H

For this cell diagram, on the left silver loses an electron to become Ag+ and subsequently reacts with iodine to form AgI. The electron passes to the cathode where chlorine ions, silver chloride, and silver are all interacting with each other.

Harrison Wang 1H

To build on this, intensive properties would be like density and melting/boiling point of an element, which remain the same despite the amount of matter present in a sample.

Harrison Wang 1H

Negative delta s does not by itself correspond to a spontaneous process, but rather the product of T and delta S should be less than delta H, which has to be negative. Negative delta S actually means that either temperature has decreased, pressure has increased, or volume has decreased.

Harrison Wang 1H

The only case where this would matter would be in a problem and some values are in j/k and kj/k. Here, you would convert all units to either j/k or kj/k.

Harrison Wang 1H

If temperature remains constant throughout the reaction, then it is said to be isothermal. In a reversible reaction, the external pressure equals the internal pressure of the container, and the external pressure is increased or decreased in small increments. In a irreversible reaction, there is a no...

Harrison Wang 1H

For a reversible process, pressure inside and outside of the system will be equal. For an irreversible process, there will be a significant difference between the pressure inside and outside of the system.

Harrison Wang 1H

It depends on the units of the constants in the problem. For example, in a problem that gives molar heat capacity, you would use kelvin, whereas in a problem that gives specific heat capacity, you would use celsius. If no preference is stated, than either or is fine.

Harrison Wang 1H

Take for example water vapor compared to water as a liquid. In water vapor, the molecules are able to move much more freely and therefore contain a larger number of states than the molecules in the liquid, which are much more constrained and limited in their movements.

Harrison Wang 1H

Heat is not an intrinsic property of a system, as it only occurs when there is a temperature difference between two objects. Intrinsic properties would be like pressure or volume.

Harrison Wang 1H

part a is -30 because the reaction is exothermic, part b is +40 because work is being done on the system, so the internal energy is 40-30=10kJ

Harrison Wang 1H

In this case, the water gives off the heat and the ice absorbs it, but yes, we always take into account the energy needed to undergo a phase change in heat equilibrium problems.

Harrison Wang 1H

We are starting from section 8.8 in the book, as everything before that deals with work and internal energy, not heat and enthalpy. Number 29 is the first question that deals with enthalpy in the book.

Harrison Wang 1H

When steam hits your hand, it first condenses, and because ∆H for the vaporization of water is extremely high, a lot of heat is released upon contact with your body. When water hits your hand, it does not require this phase change, and the heat transferred to your body is therefore much less.

Harrison Wang 1H

To find the heat needed for a temperature change, the formula is q = mC∆t, where m is mass and C is the specific heat of the substance in J/(g)(c).
To find the heat needed for a phase change, the formula is q = n∆H, where n is the number of moles and ∆H is given in kJ/mol.

Harrison Wang 1H

Ka is affected by temperature and therefore would change if it not at 25 degrees Celsius. If this is the case, I would assume the question would provide the new Ka for the specified temperature.

Harrison Wang 1H

First, you have to figure out the oxidation states of each species in each compound for the reactants and products. For the reactants, in NH4I: NH4 is +1 and I is -1 and in KNH2: K is +1 and NH2 is -1. For the products, in KI: K is +1 and I is -1, and NH3 is netural. Because K and I did not change o...

Harrison Wang 1H

A square planar shape has six areas of electron density, 2 of which are lone pairs, whereas a tetrahedal shape has four areas of electron density and no lone pairs.

Harrison Wang 1H

I think we always use ferrate for coordination compounds.

Harrison Wang 1H

An example of a T-shaped molecule would be BF3 or ClF3.

Harrison Wang 1H

No, in the VSEPR model single, double, and triple bonds all count as one area of electron density.

Harrison Wang 1H

Up to 6 ligands, which are also Lewis bases, donate their pair of electrons to a transition metal to form the cage-like molecule

Harrison Wang 1H

According to the principal quantum number (n) of period 3, the angular momentum number (l) can be 2, 1, or 0. Therefore, elements in the third period have empty d shells even though they come before elements with electrons in them already on the periodic table

Harrison Wang 1H

The first bond that atoms form is always a sigma bond, any bond after that is a pi bond, so CO has 1 sigma bond and 2 pi bonds

Harrison Wang 1H

And because molecules with pi bonds are unable to rotate, all of the atoms within the molecule are located in the same plane

Harrison Wang 1H

You should email your TA, telling him where you think you deserved more points. On past midterms, one point was deducted for incorrect units or sig figs.

Harrison Wang 1H

We are expected to know the bond angles for every structure that we have learned, this includes

Linear (180 degrees)
Trigonal planar (120)
Tetrahedral (109)
Trigonal Pyramidal (<109)
Bent (<109)
Trigonal Bipyramidal (120 and 90)
Seesaw (<120 and <90)
Octahedral (90)
Square Planar (90)

Harrison Wang 1H

The first four elements of the periodic table up to boron never obtain an octet of electrons

Harrison Wang 1H

Yes, everything up to vsper models is on the midterm

Harrison Wang 1H

The ionization energy of oxygen is higher than that of nitrogen because nitrogen has half asymmetry in its p shell while oxygen has a full first orbital. The higher electron affinity of oxygen compared to nitrogen compensates for this deficiency when talking about electronegativity.

Harrison Wang 1H

Yes, ionization energy generally increases from left to right across a period because the electrons are closer to the nucleus from the added protons and higher positive charge

Harrison Wang 1H

A quick google search turned out that molybdenum has the same valence electron configuration as chromium but tungsten does not, while both silver and gold have the same valence electron configuration as copper.

Harrison Wang 1H

Yes, the electron moves from a lower energy state to a higher energy state, which is a greater distance from the nucleus. In this case, it moves from the first shell to the second shell.

Harrison Wang 1H

^ yes, things in nature will always go to their lowest state of energy, and in this case that state is n=1

Harrison Wang 1H

To build on what Andrea said, raising the intensity of light just increased the number of photons, and without sufficient frequency these photons were still unable to eject electrons.

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