CHEMISTRY COMMUNITY

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Why is the slope of a zero order reaction always negative compared to a first or second order reaction which are negative and positive respectively?

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For something to be thermodynamically stable, it has to be spontaneous right? Even though it will not react I think for something to be thermodynamically stable, it is non spontaneous since it is unlikely to move to a higher energy on its own, making it stable thermodynamically. On the other hand, ...

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Since C is a zero order reaction, you can ignore it when solving for A and B because that portion would just cancel out to 1/1 since they are both to the zeroth power.

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does anyone know if we will be going over collision theory this upcoming week during lecture?

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Since the graph is a curve, taking the average rate will treat the graph as a line instead and not account for the curve which an instantaneous rate does.

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For problem 6O.1 we chose the cathode to be the nickel reaction which had a smaller or more negative E and this makes sense because the reaction is electrolyzed which means that a non spontaneous reaction was driven by an outside current allowing for the cathode to have a smaller E than that of the ...

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when a problem states that a reaction is electrolyzed, can we assume that the reaction is non spontaneous?

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For the maximum work of this reaction, why is the answer shown to be positive? Shouldn't it be negative because the equation for Wmax= -nFE and E needs to be positive in this case in order for the cell to do work? Thanks in advance!

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what determines whether something is a concentration cell or not? for example, in 6N3, how do we determine which reactions are concentration cells? Thanks in advance!

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How do you know when to apply Pt(S) as the inert conductor in a cell diagram besides the fact that you have elements in the same phase? Thanks in advance!

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How do you find n? Is it number of moles of products minus reactants? vice versa? or is it just the change in moles which means it is always positive? thanks in advance!

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Can someone explain why we can write Cl2 -> 2 Cl- please?

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A half reaction is balanced, when the charges on both sides of the half reaction are balanced and also when the elements themselves are balanced.

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If you figure out the oxidation numbers of each molecule, the group where the oxidation number decreases is the portion of the redox reaction that is reduced since it is gaining a more negative charge, meaning it is gaining electrons.

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If you figure out the oxidation numbers of each molecule, the group where the oxidation number increases is the portion of the redox reaction that is oxidized since it is gaining a more positive charge, meaning it is donating its electrons.

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Because the direction of flow is always from anode (negative charge) to cathode (positive charge) and the anode is always on the left while the cathode is on the right, the direction that the electron flows is always from left to right.

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jisulee1C wrote:Gibbs Free Energy. If delta G is negative you know that the forward reaction is spontaneous.

Why is it Gibbs Free Energy vs, using enthalpy or entropy.

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The Van't Hoff equation allows us to relate the equilibrium constant with temperature and also change in enthalpy and entropy.

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Yes that would make sense since in a oxidation reaction, you are losing electrons which would now be written separately in the products side of the equation.

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At equilibrium, delta G is 0 because the equation for it is deltaG= -RTlnK, and if K is at equilibrium, then K =1 and taking the natural log of 1 would give you zero.

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Just to clarify, in class Lavelle used the example:
8H+ + MnO4- + 5e- = Mn+2 + 4H2O where he said Mn is reduced from +7 to +2

Mn is initially +7 because O4 is is -8 and in order for the overall molecule to have a negative charge, Mn has to be +7 right? Thanks in advance!

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For reversible reactions, the delta s system is always equivalent to negative delta s surroundings which is equal to delta s system. However, for irreversible reactions, is delta s surroundings always equal to 0? Why? Thanks in advance!

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Increasing the pressure will shift the reaction to the side with less moles of gas.

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The answer is negative because the question is asking for the standard entropy change of the surroundings specifically and not the system so the standard entropy change of the system should be equivalent to the negative standard entropy change using the idea that when a system loses, its surrounding...

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Entropy is an extensive property because it depends on the quantity of each. For example, if you have a 6 atom molecule and a 3 atom molecule, the 6 atom molecule has a higher entropy since it can occupy more states than the latter. So in this case, since the entropy of the two molecules was depende...

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For a reversible system, the system is isothermal and thus temperature is constant which impacts the formulas utilized since some are used for temperature change and some are used when temperature is constant which is the case of a isothermal, reversible system.

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Calculate the molar kinetic energy (in joules per mole) of Kr(g) at (a) 55.85 8C and (b) 54.85 8C.

What equation are we supposed to use to start this problem?

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Calculate the work that must be done against the atmo- sphere for the expansion of the gaseous products in the combus- tion of 1.00 mol C6H6(l) at 25 8C and 1.00 bar

Can anyone give me insight on how to start this problem?

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When given the term, "heat transfer" is this referring to q or H?

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The first step is to find the heat capacity that is calibrated by the calorimeter. To do this use the eqaution C=q(calorimeter)/change in T. Essentially you would do C=(3.50kJ)/(7.32K) = 0.478kJ/K. Then you can use this heat capacity to calculate the heat released by the reaction. So set up the equ...

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Can someone explain how the equation for work for a reversible expansion is derived? or be more specific when you say the integral is taken? Thanks!

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Hess's law states that the total enthalpy change for a reaction is the sum of all changes regardless of however many steps a reaction has because enthalpy is a state property. It is typically used when you are given two or more reactions along with their corresponding enthalpy changes and asked to c...

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Bonds that are formed are negative in enthalpy calculations because energy is released when bonds are formed and the system loses energy.

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You add the q of the water and the kettle because the water in the kettle is part of the entire system now and in order to raise the kettle to the boiling point of water, the water inside needs to be raised too and thus you add the two q's.

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If heat is flowing out, the reaction is exothermic, and if heat is being absorbed, the reaction is endothermic. A trick I like to use to remember this is by looking at the first two letters of each word, in "exothermic" heat is "EXiting" and in "endothermic" heat is &qu...

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I like to think of it as the system relieving itself of the increase in pressure in order to balance itself out or reach equilibrium again by shifting to the side with less moles of gas.

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This calculate change in volume because the negative value means the volume is decreasing and also the height used which should be 20 cm is how much the pump was depressed which is then multiplied to the area of the base. The reason why this is calculating the change in volume is because the pump is...

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You would use the equation for work where W (work) = - P (work constant) x change in V. In this case you would need to calculate the change in volume first with the information given and following that, multiply it by 2.00 atm which is the work constant in this case. Your change in V should be negat...

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I believe it is used to account for the phase change and bond enthalpy when calculating overall reaction enthalpy.

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Why does the reaction shift to the side with less moles of gas when pressure increases? And why does the reaction shift to the side with more moles of gas when pressure decreases?

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When do we know when to use Ka and Kb when given a reaction equation?

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K and Kc are the same thing, they both are equilibrium constants, like the last person said, Kc just specifies that it is the constant for concentration while on the other hand, Kp, is the constant for pressure at equilibrium.

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I am still confused on where the 1/6.1*10^-3 comes from because I get that the table gives us the reverse reaction of what we want and in order to find the Kc of this reaction, we have to take the reciprocal of it. However, according to the table, the Kc for the reverse reaction is 6.1x10^23 and if ...

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Does anyone know what the very last column is used for?

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I believe that if K< 10^-3 than the reaction STRONGLY favors reactants, but if it is greater than this cutoff, it still favors reactants but not strongly? The same goes for if K> 10^3, the reaction strongly favors products but if under the cutoff, then it still favors products but not strongly.

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Ice tables demonstrate the initial concentrations, change in concentration and equilibrium concentration. So yes, by using an ice table you can not only determine the difference in concentrations but can also calculate the change.

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The reason why the reaction moves forward when the reaction quotient (Q) is less than the equilibrium constant (K) is because the lesser value means that there is less product than there should be when the reaction reaches equilibrium and thus more product is needed, so the reaction moves forward. A...

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For asking a question on chemistry community, just select the "New Topic" tab in the top left corner. First go back to the main Chem 14B page, from there you should be able to select a topic in the tabs below, and when you do, the "New Topic" button in the corner should appear. H...

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The value of the equilibrium constant when calculating using concentration versus pressure should be the same. I believe the constant only changes if you alter the temperature of the reaction but when alternating between concentration and pressure, the constant should remain the same.

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