15

Now that you've got $K_{eq}$, you need to take a look at the expression for the equilibrium constant, which is $$K_{eq} = \frac{a_{\text{ice}}}{a_{\text{water}}},$$ where $a$ is the activity of each species. For a pure liquid and a pure solid, the activity is defined as 1, so the expression on the right hand side of that equation (usually referred to as the ...


15

For a chemical reaction, $$\Delta G = -R T \ln K_{eq}$$ $$\implies K_{eq} = e^{\frac{-\Delta G}{RT}} $$ Thus, in order for a chemical reaction to have $K_{eq}=1$*, it is necessary to have $\Delta G = 0$ For some chemical reactions**, we can achieve this by adjusting the temperature and/or pressure. Thus it is possible. However, it's unlikely that the ...


9

Actually you need concept of coupled differential equations but I suppose you can do a 'trick' to avoid it: $$ \frac{dA}{dt} =- a A + b B$$ $$ - \frac{dB}{dt} = - aA + bB$$ Are the form of the two equations, you may subtract the two equation to find $ \frac{dA}{dt} = - \frac{dB}{dt}$, and by integration: $ A(t)+B(t)=C$, setting $t=0$ , we find $A(0)+B(0)=2=...


8

Your textbook's text is correct but the reaction image is incomplete. Reaction of chlorobenzene with ammonia in presence of both the cuprous chloride and cuprous oxide forms aniline but via two different reactions. Your textbook only mentions one of the reactions in the image. In the reaction involving the oxide, cuprous oxide is a reactant and is consumed ...


7

I will start by addressing the posted question: the "state of a substance at a specific temperature and pressure" refers to "the most stable phase of the homogeneous substance at the specific p and T". The phase is indicated in the phase diagram by looking up the appropriate T,p point. At $\pu{20 ^\circ C}$ and 1 atm pressure that would ...


5

[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 ...


5

$\ce{HO2-}$ is a weaker base than $\ce{NaOH}$. but $\ce{O2^2-}$ is much stronger base than $\ce{HO2-}$ and does not occur in water solutions in significant amount. But its salts ( sometimes in the form of hydrates ) can be precipitated at highly alkaline solutions of hydrogen peroxide. Additionally, lack of product presence supports the respective ...


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 ...


5

The rate forwards and back are the same at equilibrium and differ before this is attained. Let the reaction be $A+B=C$. If each concentration is suddenly halved then the reaction is not necessarily any longer at equilibrium and in this case the rates forwards and back differ until a new equilibrium is reached. The figure shows this. The red curve is the ...


5

We all are familiar with the standard state as 1 bar pressure and 273 K.The standard state can be any well defined state. You can define it in any way that is convenient for you. There is no universal definition. We may want to choose different references depending on the calculation we are doing, so we can choose different standard states. In thermodynamics ...


5

Chemical reactions may be performed in sequences (equally known as batches), or continuously. The two differ e.g., by residence time of the reagents and products in this container. The former form is like what you are used to, e.g., in the kitchen. You charge your pot with potatoes, close it, cook them, open it; done. In a flow reactor (the later form), ...


4

Note that all salts, used in humidity calibration, dissociate to their respective hydrated cations and anions when dissolved. From the solution crystallizes the sesquihydrate $\ce{K2CO3·​\frac 32 H2O}$ ("potash hydrate"). Heating this solid above 200 °C gives the anhydrous salt (Wikipedia). It seems you have anhydrous carbonate, that hydrates ...


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 ...


4

Let's say you have a reaction that has reached equilibrium. The reaction quotient $Q$ will be equal to the equilibrium constant $K$, and the Gibbs energy of reaction will be zero. Now, you decide to change the standard state (without touching your experiment, which is still at equilibrium). For example, if you are a biochemist, you could say that the pH at ...


4

Yes. The universal condition for equilibrium is that, given the current constraints on the system, no further spontaneous change can take place. I.e., that the entropy of the universe (system + surroundings) is maximized. At constant temperature and pressure, a maximization of the entropy of the universe corresponds to a minimization of the Gibbs free ...


4

First of all you made a slight mistake one of the equations is wrong First equilibrium reaction, $$\ce{Ag(aq)+ + e- <=> Ag(s)}$$ Writing the Nernst equation for this reaction $$\ce{E_{Ag|Ag^+} = E_{Ag|Ag^+}^0 - \frac{RT}{F}ln\frac{\ce{[Ag]}}{\ce[Ag^+]}}$$ Second equilibrium reaction, $$\ce{AgCl(s) + e- <=> Ag(s) + Cl-(aq)}$$ Writing the Nernst ...


4

The equation $3$ is correct because in the reaction, one gaseous molecule gets dissociated and produces a total of $2$ gaseous molecules. The number of gaseous molecules increases if the dissociation reaction proceeds. At a given pressure, the volume occupied by $2$ gaseous molecules is twice the volume occupied by $1$ gaseous molecule


4

It is the question of equilibrium versus nonequilibrium yield and production rate. Maximizing the ammonia output flow is not the same as maximizing the reaction yield. Approaching the equilibrium yield would lead to just minimal production. OTOH, the top production rate would be near zero reaction yield and unlimited input flow, what is not technically ...


3

Nothing would happen. The reaction would not shift to produce more product. Let's suppose you allow this reaction to equilibrate in a sealed container. Equilibrate means you've reached the equilibrium value for $p_{\text{O}_2}$ at that temperature. Now further suppose you were able to remove all the product without changing the free volume (say you ...


3

I am using a saturated sodium chloride solution as an example here because it is more common and easier to reproduce at home. This system is at equilibrium, with salt dissolving and crystallizing at the same rate. If you take away the solid NaCl from a saturated solution, both the forward and the reverse reactions stop. Obviously, salt crystals are no longer ...


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_\...


3

I'd like first to answer your question: My question is, when writing the formula for $K_\mathrm{D}$, is the organic phase always in the numerator and the aqueous phase in the denominator? The answer is yes. The IUPAC Recommendations 1993 (Ref.1) defines Partition Ratio ($K_\mathrm{D}$) as follows (also see the Goldbook): Partition Ratio ($K_\mathrm{D}$): ...


3

Esterification $\ce{R_1-COOH + HO-R_2 <=> R1-COO-R2 + H2O}$ has the equilibrium constant, expressed in compound activities: $$K = \frac{a_\mathrm{ester} \cdot a_\mathrm{\ce{H2O}} }{ a_\mathrm{acid} \cdot a_\mathrm{alkohol}}$$ The water activity is decreased by dissolved salts by 2 ways: decreasing the molar fraction of water by dissolved salt ...


3

You don't need gas laws, temperature or $K_p.$ You overcomplicated the problem: $K_c$ can be found by following its definition with an RICE table: $$ \begin{array}{lccccc} &\text{R}\ce{&2 HI(g) &<=> &H2(g) &+ &I2(g)} \\ &\text{I} & 2c_0 && 0 && 0 \\ &\text{C} & -2\alpha c_0 && \alpha c_0 &...


3

Option C is the right answer Solubility of AgCN will be equal to sum of concentrations of Ag+ and [Ag(CN)2]-


3

$K_p$ 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 $\...


3

Initially in the solution the following equilibria exist $$\ce{HA <=> H+ +A-}$$ and $$\ce{H2O <=> H+ +OH-}$$ Let initial concentration of $\ce{HA}$ be $a_1$ and that of $\ce{A-}$ be $a_2$. Now to a litre of this buffer solution, let's say you have added $\mathrm dn$ moles of strong acid to it. The acid dissociation constant of $\ce{HA}$ be $K_a$ ...


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} & ...


3

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}...


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