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At equilibrium, no direction is favoured. That is the point of equilibrium. If any of direction were favoured, it would not be an equilibrium. The direction with lower activation energy is probably thermodynamically favoured in context of the equilibrium constant, as this direction has negative reaction enthalpy. I say probably, as the equilibrium constant ...


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Draining the aqueous phase cannot change anything. If the phases are in equillibrium before (both internally and with each other), they stay that way if they are separated. Now if you add water to the aqueous phase, the reaction in the unpolar phase (which is saturated with water, as the phases are in equillibrium) is in the first moment not affected. ...


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NH4CN, in fact, does not dissociate completely! We write c below the respective dissociated ions to represent the concentrations at t=0 and t=teq (time at equilibrium) if there was complete dissociation. You should take in count that h stands for the degree of hydrolysis, that is, how much salt would undergo hydrolysis. Answering your second question, we ...


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Reactions that are in equilibrium will have both reactants and products present. However, the interesting thing about dissolution reactions is that a solution that is undersaturated (below its solubility limit) will not be in equilibrium (i.e. no solid exists). So unless the example you are talking about specified that the concentration of salt is above the ...


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Your question and your formulas are really asking about two different things. Let's look at your formulas first. $$\ce{H2O + HA <=> H3O+ + A-},\tag{R1}$$ $$\ce{HA <=> H+ + A-}?\tag{R2}$$ For R1 the equilibrium equation should be: $$\mathrm{K_a} = \dfrac{\ce{[H3O+][A-]}}{\ce{[H2O][HA]}}\tag{Eq-1}$$ but for R2 the equilibrium equation should be: $$\...


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The equation $\ce{N_2O_5 <=> N_2O_3 + O_2}$ is really two equations $\ce{N_2O_5 \to N_2O_3 + O_2}$ and the reverse $\ce{ N_2O_3 + O_2 \to N_2O_5}$. Suppose we start with pure $\ce{N_2O5}$ frozen at 77K so no reaction can occur then suddenly heat it up to room temperature so that reaction starts. As soon as $\ce{N_2O3}$ is formed there is the ...


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"In this case, the equilibrium constant suggests the products are favoured but if you look at how much product we have compared to reactants there's so much more reactants than products." This is what I take as the heart of your question. To that point: I believe you need to compare the concentration of all the reactantS, not just one of the ...


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Let's make this an abstract problem with species A, B, C, and F: $$\ce{A <=> B + F}\tag{1}$$ $$\ce{B <=> C + F}\tag{2}$$ We can also write down the sum of the two reactions, which will also be at equilibrium: $$\ce{A <=> C + 2 F}\tag{3 = 1+2}$$ And let's use a for the concentration of A and so on. Finding the solution using reactions 1 and ...


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METHOD 1 Suppose we were to start off with pure reactants in stoichiometric proportions in the standard state of 1 bar, and end up with pure products in corresponding proportions in the standard state of 1 bar. The change in Gibbs free energy for this process is $\Delta G^0$, as you have answered in your comment to me. To bring about this change, we use ...


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For an isobaric process you can integrate the expression $dS = dq/T = C_p dT/T$ to compute the entropy change at another temperature. Since the heat capacities are assumed constant you can proceed as follows: $$\begin{align}\Delta S_{\pu{380 K}}^\circ &= \Delta S_{\pu{298.15 K}}^\circ + \sum_i \nu_i\int_{298.15 K}^\pu{380 K} \frac{C_{p,i}}{T}dT\\&= \...


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There are two different answers, depending on whether you mean that the alternate pathway can occur in the same reaction mix (ie under the same conditions) as the described forward reaction is occurring. If there is an alternate pathway possible under the same reaction conditions, it will be used for both the forward and reverse reactions. Due to the ...


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Let's start from the equilibrium constant, which may be defined $\ce{K=\frac{[C][D]}{[A][B]}}$. Of course, activities should be introduced here instead of concentration. But it is not the point here. The main use of this constant is to calculate how the composition of the system changes if at least one of the concentrations changes. But if one of these ...


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Rather, active mass of solids need not necessarily be unity, but arbitrarily is unity with advantage. You do not have to know the molar mass nor density. The common thermodynamic expressions like $K = \frac{[C][D]}{[A][B]}$ or kinetic ones like $\frac{\mathrm{d}[C]}{\mathrm{d}t} = k \cdot [A] \cdot [B]$ should be more exactly written as $K = \frac{a_\mathrm{...


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Yes, you can. An ideal Carnot cycle (admittedly not a reaction) illustrates how this is possible. In such a cycle each step is reversible. To return to a state you can either reverse the direction of the process (running say a heat engine as a refrigerator) or follow the full path of the cycle until the state is again encountered.


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Here is the expression from the question: $\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} = K_{eq}$ Get rid of the fraction: $(A_0 - [C]_{eq})(B_0 - [C]_{eq}) = K_{eq} {[C]_{eq}}$ Distribute the sums in the product: $A_0 B_0 - [C]_{eq}(A_0 + B_0) + [C]_{eq}^2 = K_{eq} {[C]_{eq}}$ Sort the terms and you have your quadratic equation: $A_0 B_0 ...


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It should be noted that this formula is applicable only if the acid constant of the weak acid is indeed much smaller than the concentration of the H3O+ ions formed in the protolyzation of the strong acid. This is because the formula uses the approximation K1 << C2 in the derivation. For example, if: $$ K_1 = 5 \cdot 10^{-2}, $$ and the $$ C_2 = 0.01 M, ...


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First of all, the indicator does not change colour instantaneously at $\mathrm{pH = pK_a}$. The indicator is a different colour than it's conjugate base, and exists in an equilibrium with it in solution. For the colour change to be noticed, the concentration of the base has to be at least 10 times more than that of the conjugate acid, or vice versa. This is ...


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The rate of a reaction, or how fast reactants become products, depends on the collisions of particles. This is because when a particle collides, the reaction occurs (provided that it has enough KE). So, if we increase the amount of particles (the concentration), we will have more collisions and therefore the reaction will go by faster.


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The answer may be somewhat counterintuitive. Yes, an inert gas does very much exert pressure on a solid or liquid. The problem is that the chemical potential of the substance in the gas phase is not affected by the inert gas (if the gas is assumed ideal). Assume you have a substance (liquid or solid) in a rigid isothermal but otherwise empty box. The result ...


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