New answers tagged physical-chemistry
0
16 g of dioxygen contains 16 g of oxygen atoms. To find out how much carbon dioxide is produced, you can either use a realistic description of the reaction, or you can go atom by atom. Either path gives the same solution, but the one chosen by the textbook is not modern chemistry, and makes some of us angry or sad.
My question is why not to use the reaction ...
0
Of course it is a pipe and it has a infinite volume hypotactically,
To use ideal gas equations is a bit tricky with mass flow rate need to convert to molar flow rate of each gas thus need to divide by $\ce{M_w}$ molecular weight respective to each gas.
Consider $\ce{1}$ time unit where inside of pipe gets
$$\frac{Q_1}{M_{w_1}}$$
$$\frac{Q_2}{M_{w_2}}$$
moles ...
0
As suggested in the comments, the key is Nernst equation. The fundation is the Gibbs free energy given by:
$\Delta G = \Delta G^0 + RT \ln Q_r$
where $\Delta G^0$ is the standard free energy for the process, while $Q_r$ is the reaction ratio. Remember that the reaction ratio is the analogous quantity of the equilibrium constant, but with the non equilibrium ...
3
You have not propagated your errors properly. Note that the sum in your nominator evaluates to a small number with an error of roughly the same size (99.73±0.01 - 99.72±0.01 = 0.01±0.01). Also, error propagation cannot be done by simply adding up the errors.
Instead of calculating the error propagation manually (cumbersome, and prone to errors, hehe ;-) I ...
0
I've outlined the derivation of the equation, a more detailed explanation can be found in many phys. chem. and thermodynamics textbooks. Suppose, for simplicity, that the equilibrium is A(g)=B(g) and at constant temperature and pressure. The Gibbs free energy change for the reaction is $dG=(\mu_B-\mu_A)d\xi$ where $\xi$ is the extent of reaction which is $0$ ...
2
$\ ΔH_{sub}(I_2,s)$ can be interpreted as the change in enthalpy when one mole of solid $\ I_2$ converts to gaseous $\ I_2$ at a constant temperature. So the answer should be reaction (b).
3
Yes, there is. The following is a plot of the hydration enthalpy versus row in the Periodic Table. Data was extracted from the Wikipedia, not the primary source (Ref 1).
Your thinking is right. The hydration entropy is evidently negative due to structuring of water (including loss of rotational degrees of freedom), becoming less negative for the larger ions (...
0
I think what OP need is the derivation of the equation $(1)$ in the question (as indicated in the OP's comment, "an anyone help me with the derivation of the first one").
The first part of this derivation, the formula for the rate constant $k$ for the first order reaction, is done in the last part of Chemistry student's answer, and hence I'd not ...
1
In order to compare $K$ and $Q$, you have to define them in terms of the same measures of concentration and the same standard state. The only difference in the definition should be that in order to determine $K$, you are required to use the concentrations or activities of an equilibrium state, while $Q$ may be determined for states that are not necessarily ...
0
Under constant pressure and temperature, we have
$$-\frac{\Delta G}{T} = \Delta S_{tot} = \Delta S_{sys} + \Delta S_{surr}$$
hence, a negative $\Delta G$ tells us that the total entropy gets increased (= spontaneous process). A common way for textbooks to justify the definition of $G$ is by specifying $\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$, which is ...
1
Standard Enthalpies of Reaction
The magnitude of $\Delta H^o_f$ is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients:
Here is the table of enthalpies required to break bounds for the ...
1
The half-life $t_½$ is the time it takes for the concentration of the first order reaction to decrease by a factor of two. If we plug this into equation (2) given by the OP, we get
$$\ln\frac{1}{2} = -k t_½$$
We solve this for $k$ to get:
$$k = \frac{ \ln(2) }{t_½} \approx \frac{0.693}{t_½} $$
My lecturer mentioned that the formula for the rate constant $k$ ...
3
I am attaching a written answer as typing the log and other exponents is hard :(
1
The two equations are pretty the same. The number 2.30 is used to convert the natural logarithm in decimal logarithm: $\ln 10 = 2.303$. Also, substitute $[A]_{1/2} = [A]_0 /2$, from the definition. The concentration at the half-life is exactly the half of the initial concentration, then you will obtain $\ln 2$ inside the logarithm term.
1
Given the whitening observed at the cathode, one potential explanation is that calcium salts were precipitating there, forming a layer blocking passage of part of the current.
I usually write the cathodic hydrogen-evolution reaction in neutral solution as:
$$
\ce{2 H2O + 2 e- -> H2 + 2OH-}
$$
Thus, the pH in the vicinity of the cathode will be ...
2
The only opportunity where the equivalents are still used today is in the measurement of pollution in used waters. As the nature of the dissolved impurities is not known, their amount cannot be precisely known or weighed. But they can be characterized by the amount of electrons consumed by a strong oxidant like $\ce{K2Cr2O7^}$ or $\ce{KMnO4}$ if this ...
3
Wow, equivalent weights keep popping up again and again. It is not a nebulous term, the point is you are checking recent webpages. You have to look up chemistry books from 1930 to 60. If corona were not an issue I would have suggested a library trip to read Vogel's Quantitative Chemical Analysis (oldest edition).
Equivalent weight has nothing to do with mol. ...
2
OP’s question 1: Where did I go wrong with that?
OP’s only mistake on the calculation here is the assumption of alcohol and water are additive. However, as matt_black pointed out in his answer that alcohol and water are not additive since the smaller water molecules can take up some of the space between the larger alcohol molecules, causing volume reduction ...
3
I would like to give an alternative derivation for theorist's answer. You can get the same answer with statistical mechanics using a lattice model.
Let us discretize the container into $M_1$ respective $M_2$ cells. The volume of a cell is $V_{cell}$. Each gas particle fits exactly into one cell and can hop from cell to cell in a discrete manner. Since we are ...
2
The solubility product of $\ce{PbSO4}$ is about $10^{-8}$. Now suppose $[\ce{Pb^{2+}}$] falls down from $\pu{1 M}$ to an arbitrary low value like $\pu{10^{-8} M},$ due to addition of $\ce{SO4^{2-}}$ ions. In this case, Nernst's law can be applied, and the potential of the lead electrode falls from $E^\circ_\ce{Pb} = \pu{-0.13 V}$ down to
$$E_\ce{Pb} = \pu{-0....
1
There's a simple way to solve this: We know that, for an ideal gas at constant T:
$$\Delta S = -nR \ln \frac{p_f}{p_i}= nR \ln \frac{p_i}{p_f}$$
And since ideal gases ignore each other, we can calculate the entropy change for gas 1 and gas 2 separately, and simply sum the two. [At any given $T$, the chemical potential of an ideal gas is determined only by ...
2
The thermodynamic data refers to formation of the given compounds from the elements in their standard states. For instance for $ \ce{ ZnO(s)}$ it refers to the reaction
$ \ce{ Zn(s) +1/2O2(g) ->ZnO(s)} \tag{1}$
with each element or compound present at a pressure of 1 bar and (unless noted otherwise) a temperature of 298.15 K.
A more negative value of $\...
2
I think you can look at this entirely from the perspective of entropy (forget about the Gibbs free energy for a moment). You can divide the universe into two parts, the system and its surroundings.
When a change happens in your system, and some heat is exchanged with the surroundings then the change in entropy of the surroundings is: $$\mathrm{\Delta S_{surr}...
0
The total differential of the internal energy (at constant composition and allowing only pV work) is usually written as
$$dU = -pdV + TdS$$
However the differential $dU$ can be written in terms of other variables, for instance p and T, in which case:
$$\begin{align} dU &= \left(\frac{\partial U}{\partial V} \right)_T dV + \left(\frac{\partial U}{\partial ...
1
For equally charged ions a smaller radius implies a higher charge density, and therefore stronger Coulombic interactions with other charges in solution, including stronger ion-dipole interactions with water molecules, resulting in a more stable hydration sphere, and stronger interactions with counterions (here chloride). Such interactions reduce the mobility ...
0
As an answer to complete the previous ones, I've got this issue many times in the past for my calculations and tried several options to have full convergence for my optimizations. As mentioned by the author, an optimization job followed by a frequency job are required to find a stationary point and check it through frequencies analysis. However sometimes, ...
0
First, read my answer here where I cite the electrochemistry background. The latter include source links to galvanic corrosion and, in particular, in the current instance, a classic case of pit corrosion.
Stop long-term soaking, as this in the presence of salts and an H+ together with dissolved oxygen will just further promote the slowly evolving ...
0
The internal energy of a closed system is $dU=TdS-PdV $. In an isolated system $dU=0, dV=0$ which would imply that $dS=0$, However, the entropy can increase due to chemical reaction or diffusion mixing different substances which were initially separate. Similarly $dG=-SdT+VdP$ and if $T$ and $P$ are held constant this does not mean that $G$ cannot change as ...
3
Note that under the given conditions, methanol cannot exist in the liquid state, however since you are given in the problem that methanol is obtained in liquid state, you will only consider the equilibrium (for problem solving purpose)
$$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)}$$
And since $K_p$ is related to $K_c$ by the relation
$$K_p = K_c(RT)^{\Delta n_\...
0
In less mathematical terms, the chemical potential of a component in a mixture at a specific temperature and pressure is the amount of free energy that can be ascribed to one mole of that component under the conditions found in the mixture at that T and p. Alternately it is how much the free energy of the system, if it were scaled to an infinitely large size,...
0
As porphyrin and Buck Thorn nicely explained, once you establish that $K_p$ is independent of pressure then, assuming an ideal system, it necessarily follows that $K_c$ is indepedent of pressure as well.
But I'd like to take a closer look at the pressure-independence of $K_p$. The reason $K_p$ is pressure-independent is not directly because it is defined as ...
0
Elaborating on @porphyrin's answer, for an ideal system $K_c$ does not depend on pressure because if the temperature is constant then the concentration of a component depends only on its partial pressure $p_i$ and not on the total pressure P:
$c_i=\frac{p_i}{RT}$
In other words, at constant T a number concentration can be used as a proxy for the partial ...
1
Let's see if we can tease out the main points here. The images below are taken from Lance and Cole's Analytical Chemistry POGIL workbook.
Microscopic processes at a metal surface exposed to an aqueous solution containing the corresponding metal ion(1) will result in a slight negative charge on the metal surface due to oxidation of the metal and a slight ...
4
By saying Catalysts are not consumed by reactions. is meant there is no stoichimetric ratio to reactants, consuming catalysts and forming from them catalytically inactive compound.
By other words, some of reaction steps regenerates the original form of a catalyst, consumed by a prior step, so the net consumption is negligible.
The transition between active ...
-1
Consider the equilibrium: A+2B (reversible reaction arrow) 2C.
If [C] and [B] are 2 moldm^-3 while [A] is 1 moldm-3 originally: Kc = [C]^2/([A][B]^2) = 2^2/(12^2) = 0.8 mol^-1dm^3
If pressure increases and nothing happens (which cannot happen in actual): [A], [B] and [C] increase by say, 3/2 times because an increase in pressure can only be due to an ...
2
To answer the general part of the question: no, we are not limited to the Cartesian planes and axes. The obvious examples are the ammonia and the methane milecules: the former has several reflection planes, positioned at $120^\circ$, whereas the latter has multiple reflection planes and also the rotation axes (one can easily google both).
Regarding the ...
3
I don't see planes of symmetry through H atoms or OH groups. Actually, these planes are not present. It is easy to check this by looking at the position of the CH3 group after reflection in one of these planes: you find CH3 at a completely different position (none of the original positions is mapped and hence you can distinguish the starting configuration ...
2
$K_p$ is defined in terms of free energies of the gaseous substances in their standard states at 1 atm pressure so must be independent of pressure, i.e. is a constant. With concentrations we use $c_i=n_i/V$ for species $i$ and for example, if $K_c=c_\ce{A}^ac_\ce{B}^b/(c_\ce{C}^cc_\ce{D}^d)$ then for an ideal/perfect gas $K_c=K_p(RT)^{-\Delta n}$ where $\...
0
The initial state of your system consists of 4 moles of $\ce{Fe(s)}$ and 3 moles of water vapor at $\pu{298 K}$ and $\ce{1 bar}$ in separate containers. The final state of your system consists of 2 moles of $\ce{Fe2O3(s)}$ and 3 moles of $\ce{H2(g)}$ at $\pu{298 K}$ and $\pu{1 bar}$ in separate containers.
The $\Delta H^\circ$ and $\Delta S^\circ$ correspond ...
3
If I understand your question, you would like to know how the Avogadro number is measured. There are no simple and immediate answer.
One of the possibility is using X-rays to measure with precision the distance $d$ between two atoms in a cubic crystal like $\ce{Si}$, where the atoms $\ce{Si}$ are regularly alined along the three axes Ox, Oy, Oz.
As the ...
0
Theoretically, we could do any reaction.
Practically, some are only possible in one direction; others are under special conditions.
For the case of copper: the standard potential of the pair $ \ce {Cu^{2 +} / Cu} $ being positive, this means that naturally the reaction is done in the direction $ \ce {Cu^{2+} -> Cu} $.
But we can still force the reaction ...
1
In the "Index of Electrolytes" on page 3 of the document at the source of the problem (Ref. 1)
note that the third column is titled "Maximum conductance and point of inflection at 25°C [$\pu{μmhos/cm}$/% by wt.]". However, the entries are given in units of conductivity ($\pu{μmhos/cm}$ or $\pu{μS/cm}$)! The first number for each entry in ...
2
I believe the vapour produced in the step 3 of you experiment is due to the particles with high order speed in the Maxwell distribution of speeds curve that escape because of their higher kinetic energies. When escaping, they will reduce the average kinetic energy of particles in the water sample, which will then reduce the net temperature of sample.
This ...
6
The answer is quite simple, if you look at the equation carefully. Conductance and conductivity are related by cell constant, $\displaystyle\frac{\text{Area}}{\text{Distance}},$ where the area is the area of the electrodes and the distance is the distance between the electrodes.
$$G = \sigma\frac{A}{l}$$
In conductivity measurements, the cell constant was ...
-1
the standard potentials are given in the literature generally in an acidic medium at pH = 0. In this case, we must therefore write the equations in an acidic medium with $ H ^ + $ and not with $ HO ^ - $
0
I think that you should focus on what you wrote first, longer the hydrocarbon chain higher their entropy in solution. When dissolved, they can access a multitude of states both energetically and in spaces (think of all possible conformers).
This said, it does not mean that they must dissolve in, eg, water. Right because the H term.
5
I'll try to give some information that would help you according to your statement in the question:
I've actually been trying to find more generalised information to help with the assignment of a different (but similar) cyclohexene compound.
First of all, you may have found now from the two answers given elsewhere that there are no way you can assign all ...
2
The equation
$$\ce{2A <=> B + C}$$
implies a fixed stoichiometry between B and C. If the two separate reactions (2) and (3) were happening, the correct net equation would be
$$\ce{(x + y) A <=> x B + y C}$$
Here is an example that would be well-described by reaction (1):
$$\ce{2 H2O <=> H3O+ + OH-}$$
One water molecule cannot make a ...
-1
You have stated that the reaction:
$$\ce{2A <=> B + C}\tag{1}$$
[...] can be thought the sum of the following reactions:
$$\ce{A <=> B}\tag{2}$$
$$\ce{A <=> C}\tag{3}$$
However, this is not always the case. In particular, your assertion that the overall reaction (1) is a linear combination of reactions (2) and (3), is based upon the ...
11
Referring to your comment on Buttonwood's answer:
Unfortunately, in this case I only have 1D HNMR and IR available to analyse the product
If this is really the case, then you are essentially out of luck. It is not possible to extract much information from a series of overlapping multiplets, which may well be coupled to each other (these strong coupling ...
Top 50 recent answers are included
