5

The essential condition is the compound thermal stability. If it decomposes below its melting point, it does not have a triple point. If it decomposes before properties of gaseous and liquid phases converge to each other, it has just estimated, extrapolated triple point. Beyond CP we talk about supercritical fluid, it kind of shares many properties of both ...


4

Recall that $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ Where $\ce{[OH-]}$ is the concentration of hydroxide ions. Suppose a solution has $10^{-7}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-7}]$ which is equal to $7$. Suppose a solution has $10^{-4}$ ...


4

An exothermic reaction has a reduced equilibrium constant at higher $T$ because while the contribution of the change in the entropy of the system is a fixed quantity (for a small $T$ change), the effect of transferring heat to the surroundings is reduced at higher $T$ (because it causes a smaller change in the entropy of the surroundings). Mathematically ...


3

[OP] The equilibrium constant would not include the solid $\ce{I2}$, but why is this? Let me explain this with a different example. If you have a saturated solution (e.g. lemonade with too much sugar in it) it is at equilibrium. If you add more sugar, the lemonade does not get sweeter. That tells you that the amount of solid does not matter (as long as ...


3

I will continue with the data of $[\ce{H2CO3}] = \pu{10^{-4.97} M}.$ Now, as the $K_2$ of $\ce{H2CO3}$ is very small as compared to its $K_1,$ we can assume that all the $\ce{H+}$ will come from the first dissociation of $\ce{H2CO3}.$ $$ \begin{array}{lccc} & \ce{&H2CO3 &<=> &H+(aq) &+ &HCO3-(aq)} \\ &\text{Initial} & ...


2

The free energy difference between reactants and products are given as follows when both are in their STANDARD states: $$\Delta G^\circ = -RT \ln K_\mathrm{eq}$$ If $K_\mathrm{eq}$ is less than one that means $\Delta G^\circ$ is positive and the reaction as written will go to the left until the equilibrium conditions are satisfied. At this time $\Delta G = 0$...


2

The $\ce{Na+}$ ion is a spectator ion in this case. It doesn't participate in the chemical reaction. It is found unchanged on both the reagent and product sides (red top-right and green bottom-right reaction formulas), so it can be canceled out to get the net reaction (red left). From there, he did the usual acid-base calculation to get a formula for $\ce{Kb}...


2

This was answered in the comments, but these second equilibrium expression should not actually be a $K_b$ value. Rather, it should be $1/K_a = 55600$. Using this value in the same calculations, we find that $x=5.5805\times 10^{-4}$ and $y = 5.5795\times 10^{-4}$, so that $\ce{pH} = -\log(x-y) = 7.00$.


2

The autoionization equilibrium of water is satisfied irrespectively of $\mathrm p\ce{H}$. That's why $\mathrm p\ce{H}$ and $\mathrm p\ce{OH}$ add up to a constant: $$K_\mathrm{w}=a_{\ce{H3O+}}a_{\ce{OH-}}$$ $$\rightarrow \mathrm p K_\mathrm{w}=\mathrm p\ce{H} + \mathrm p\ce{OH}$$ This is a bit different from the typical scenario in which Le Chaterlier's ...


2

Question: Determine the $\mathrm{pH}$ of the solution resulting when $\pu{100 cm^3}$ of $\pu{0.50 mol dm-3}$ $\ce{CH2ClCOOH}$ is mixed with $\pu{200 cm^3}$ of $\pu{0.10 mol dm-3}$ $\ce{NaOH}$. I'm not sure what level of chemistry is OP's in, but the given solution for the question is for the chemistry students with appreciable knowledge of stoichiometry of ...


1

The graph is a little bit confused, because there exist different sorts of Gibbs energy. i It can be the Gibbs energy of the rectant. It can be the Gibbs energy of the products. And it can also be the total of these two Gibbs energies. First let's discuss the Gibbs energy of the reactant. In the beginning of the reaction, the Gibbs energy of the reactant is ...


1

Here is a labeled and annotated version of your graph (from OpenStax Chemistry).


1

An analogy might help. Solubilization of a salt in a solvent (say NaCl in water) is similar to condensation of a gas. Imagine you have an amount of gas in a cylinder with a piston to regulate the volume and pressure. Beginning at a pressure below the vapor pressure of the substance (which means that all is gas, none is liquid), the pressure is increased (...


1

$\ce{CH2ClCOOH}$ is a weak acid; $\ce{NaOH}$ is a strong base. Now, tell me, what do you think happens when you mix a strong base with a weak acid? They react. $$\ce{CH2ClCOOH + NaOH -> CH2ClCOONa + H2O} $$ A neutralisation reaction in aqueous medium depends upon the number of equivalents of acid and base. $$N_1V_1=N_2V_2 $$ At this stage, an equivalence ...


1

Why should they ( supposing ideal gas behaviour ) ? If a volume, containing $\pu{4 mol}$ $\ce{N2}$ and $\pu{1 mol}$ $\ce{O2}$ has pressure $\pu{5 atm}$, partial pressures are: $\ce{N2} : \pu{4 atm}$ $\ce{O2} : \pu{1 atm}$ If you add $\pu{5 mol}$ $\ce{Ar}$, the total pressure raises to $\pu{10 atm}$, with partial pressures : $\ce{Ar} : \pu{5 atm}$ $\ce{N2} : ...


1

You must first understand what happens when water is heated along the boiling curve. Start from the point $\pu{100°C}$and $\pu{1 atm}$. At this point there are two phases, one liquid whose density is nearly $\pu{1 g/cm^3}$ and exactly $\pu{0.96 g/cm3}$. And the vapor has a density of $\ce{0.0006 g/cm^3}$. If now you heat this system in a closed volume, the ...


1

Acids are hydrogen ion donors. When the react with water, they can give a hydrogen ion to form $ \ce{H3O+}$. For example: $ \ce{HCl(aq) + H2O(l) \rightarrow Cl-(aq) + H3O+(aq)}$ Simple acids, such as $ \ce{HCl}$ or $\ce{H2SO4}$, can be recognized as acids by the H at the start of the formula. Other more complex acids may be written with $ \ce{COOH}$ at ...


1

H3O+ is just the combination of a a $\ce{H+}$ ion, which we know to be released from the dissociation of an acid in an aqueous solution, with a water molecule. It happens because the $\ce{H+}$ is so, so positive (and therefore so reactive) that the water molecules, with its lone pairs (which are locally negatively charged), are willing to form a dative ...


1

I know you wanted something non-mathematical, but I think the best way to understand this intutively requires some simple math. Suppose you have the following elementary reaction: $$\ce{A(g) + B(g) <=> C(g)}$$ By "elementary reaction" I mean that this shows the actual reaction mechanism, such that the rate equation can be obtained directly ...


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