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The physical reason why the dimensions of k differ for different reactions is the different mechanism behind the reaction. A process that only involves the internal vibration or rearrangement of one species differs from a process in which two or more equal or different species must first diffuse and collide*. It seems intuitive to me that this should be ...


4

Maybe it would make more sense if rate constants would have different symbols depending on whether it is a zero, first or second order reaction. As an analogy, we might be asking "how big" an object is depending on the length of it. Depending whether the object is, say, a square, a cube or a hypercube, the answer would be in terms of its area, ...


4

The solubility of $\ce{AgCl}$ is equal to $\sqrt{K_\mathrm{s}} = \pu{1.3E-5 M}.$ The solubility $s$ of $\ce{Ag2CO3}$ is such that $K_\mathrm{s} = 4s^3.$ So that its solubility $s$ is equal to $s = \pu{1.2E-4 M}.$ This is ten times more than the solubility of $\ce{AgCl}.$ For gravimetric purposes, $\ce{AgCl}$ is a better choice.


3

The above simplified formula considers 2 simplifying assumptions: Concentration of hydrogen ions from water autodissociation can be neglected, so $[\ce{H+}] \approx [\ce{A-}]$ . Concentration of an acid $c_0$ , in relation to its $\mathrm{p}K_\mathrm{a}$, is high enough for the fraction of ionized acid to be negligible, so $c_0 \approx [\ce{HA}]$ . ...


3

Maybe a numerical example may help you compare the different orders. Let's take first an example of a reaction of zeroth order : the combustion in air of a candle containing $n$ $mol$ wax. At every second, the same amount of wax is burned. The rate of the reaction is $$r = -\frac {dn}{dt} = k_o·n^0 = k_o$$ If the candle contains $1.44$ mole wax and burns in $...


2

Well, it may not be appropriate to say that the rate constant, k, has no physical meaning. It is the only factor in the rate equation which is temperature dependent, so larger k means faster rate of reaction. Another way of looking at the units of k is in terms of Arrhenius equation. $$ k(T)=A e^{-E_{\mathrm{a}} / R T} $$ The factor $A$ is called the ...


2

The reason is that strong acids have $\mathrm{p}K_\mathrm{a}$ values that are poorly known. These $\mathrm{p}K_\mathrm{a}$s cannot be determined directly, because the exact concentration of the ions is difficult to know with precision: the electrodes do not react with the concentration of $\ce{H^+}$ or $\ce{H3O^+}$ ions. They react with the activity of $\ce{...


2

I think your thought process is correct, but you can push it further. The way I would solve this problem is to think about how you could go from the alkyne to either alkene. Then, via microscopic reversibility, you would argue that the reverse mechanism--the elimination--is very similar.


2

The simplest way to approach this is that the rate constant has whatever units that make the rate have the correct units. Generally, the rate is in units of molar per second, so based on how you're multiplying concentrations (in molar) in the rate law, you should be able to figure out the appropriate units for the rate constant.


2

In a weak acid $\ce{HB}$ solution, with a nominal concentration $c$, a tiny amount of its molecules are dissociated into $\ce{H^+}$ and $\ce{B^-}$. Let's call this concentration $[\ce{H^+}]$ = $\ce{[B^-]}$ = $x <<c$, so that the following approximation can be made : $c - x$= $c$. The dissociation equilibrium constant $K_a$ of this weak acid $\ce{HB}$ ...


1

You have, for each compound, $\mu_i = \mu^\circ_i + RT\ln(P_i/P^\circ)$. So, replacing each potential in $\Delta G$ by its expression, assuming $P^\circ = 1$ bar, we get : $$\Delta G = c(\mu^\circ_C + RT\ln(P_C)) + d(\mu^\circ_D + RT\ln(P_D)) - a((\mu^\circ_A + RT\ln(P_A)) - b(\mu^\circ_B + RT\ln(P_B))$$ Then we develop and group standard potentials and logs:...


1

Be aware that in the solubility comparison context, solubility product constants can be directly compared for compounds with the same number of ions created from the formula, where a greater solubility product means a greater solubility. For compounds with different ion counts, one has to compare ( molar ) solubilities in $\pu{[mol/L]}$ , calculated from ...


1

Both water samples ( assuming pure water ) would theoretically end up slightly acidic with $\mathrm{pH} \approx 5.6$, but the carbonated one after long time. Tap water would have its baseline $\mathrm{pH}$ determined by its $\ce{CO2(aq)/HCO3-(aq)}$ $\mathrm{pH}$ buffer. This is valid for an ideal case, as side processes - especially for tap water - will ...


1

All reactants and products are in the liquid phase and mixed together in one vessel, so the total volume at equilibrium, $V$, is the same for each concentration we need to find $K$. Also, all reactants and products are in one to one molar ratio in the balanced chemical equation. Therefore, if we start with $\pu{3 mol}$ of each reactant, and reach ...


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