New answers tagged equilibrium
-1
As Chet Miller and Karsten pointed out, the culprit is the simple definition of K which we use in textbooks.
As in your last equation $$Gi(Pi)=G(P∘)+niRTln(Pi/P∘)$$
(similar equation can be used for dilute solutions in liquid phase)
This equation (differential of this equation in the form of chemical potential) is the origin point of the relation $$ \Delta G ...
3
The proton transfers may happen in cascade.
For example, pure $\ce{HCl}$ is a gas that dissolves and reacts with water (acting as a base) according to a proton transfer from HCl to $\ce{H2O}$ : $$\ce{HCl + H2O -> H3O+ + Cl- ... \ (1)}$$ But $\ce{H2O}$ does not hold its proton very strongly. Here is why : if another base like ammonia is added to this ...
2
Let's say you have a reaction that has reached equilibrium. The reaction quotient $Q$ will be equal to the equilibrium constant $K$, and the Gibbs energy of reaction will be zero. Now, you decide to change the standard state (without touching your experiment, which is still at equilibrium). For example, if you are a biochemist, you could say that the pH at ...
2
We all are familiar with the standard state as 1 bar pressure and 273 K.The standard state can be any well defined state. You can define it in any way that is convenient for you. There is no universal definition. We may want to choose different references depending on the calculation we are doing, so we can choose different standard states.
In thermodynamics ...
2
Initially in the solution the following equilibria exist
$$\ce{HA <=> H+ +A-}$$
and
$$\ce{H2O <=> H+ +OH-}$$
Let initial concentration of $\ce{HA}$ be $a_1$ and that of $\ce{A-}$ be $a_2$. Now to a litre of this buffer solution, let's say you have added $\mathrm dn$ moles of strong acid to it. The acid dissociation constant of $\ce{HA}$ be $K_a$ ...
2
One of the first experimental demonstrations that the same equilibrium point is reached if the reaction starts with all reactants or all products was $\ce{2HI<=>I_2 + H_2}$ made by Bodenstein in $\approx 1894$. I have redrawn the data.
figure. $\ce{ 2HI=I_2 + H_2}$. $x$ is the fraction of HI at $721$ K. $x=[HI]/([HI]+2[I_2])$
In the reaction $\ce{A&...
0
If the relative humidity is $80$% and the saturation water vapor pressure is $\pu{85 mmHg}$ at $\pu{300 K}$, it means that the vapor pressure in the flask is $\pu{0.8 · 85 = 68 mmHg}$. We will now show that this value is much higher than the value coming from equilibrium constant of the reaction. As this constant is : $\pu{K_p = 400 mmHg^2}$, the value of ...
3
Suppose initially you had 1 mole of $\ce{N2}$ and 3 moles of $\ce{H2}$, and that $n$ moles of $\ce{N2}$ had reacted. Then you would have, at equilibrium, $1-n$ moles of $\ce{N2}$, $3(1-n)$ moles of $\ce{H2}$, and $2n$ moles of $\ce{NH_3}$. So the total moles at equilibrium would be $4-2n$, and the mole fractions (equal to the partial pressures in atm) ...
14
For a chemical reaction,
$$\Delta G = -R T \ln K_{eq}$$
$$\implies K_{eq} = e^{\frac{-\Delta G}{RT}} $$
Thus, in order for a chemical reaction to have $K_{eq}=1$*, it is necessary to have $\Delta G = 0$
For some chemical reactions**, we can achieve this by adjusting the temperature and/or pressure. Thus it is possible. However, it's unlikely that the ...
0
The apparent contradiction comes from the fact that in the case of Atkins' Physical Chemistry there is no assumption of equilibrium, just that the surroundings have a temperature that is not affected by whatever happens to the system.
2
As usual, I suggest to start with a RICE table to get an overview, denoting initial partial pressures with $p_0$ and the change in partial pressure with $x$:
$$
\begin{array}{lccccc}
&\text{R} &\ce{&CH3OH(g) &+ &NOCl(g) &<=> &CH3ONO(g) &+ &HCl(g)}\\
&\text{I} && p_0(\ce{CH3OH}) && p_0(\ce{NOCl}) &...
2
When you consider the dissolution of a gas like $\ce{SO3}$ , $\ce{CO2}$, $\ce{SO2}$ etc. there are three equilibria existing at the same time:
$$\ce{SO2(g)\overset{H2O}{<=>}SO2(aq)\overset{H2O}{<=>}H2SO3(aq)<=>H+(aq) +HSO3-(aq)}$$
The first equilibrium is the dissolution equilibrium of the gas, the second is the hydration equilibrium of the ...
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 &...
2
When ionic compounds dissolve in water, water molecules form a complex ion around the calcium cation. The presence of the hydroxide anion has a similar effect, though the polarities of the water ligands is reversed.
Thus, dissolution of Ca(OH)2 in water is actually a chemical reaction. In some cases, complex ions may be more structured than the crystals of ...
-2
Entropy can be measured and aproached differently, (i) from an equation that relates entropy and Enthalpy to measure the heat of reaction, the remainder of the work done per degree of Temperature (°) is
or if the change in enthalpy is measured then simply.
In this case the Vector of entropy has a negative direction hence an endothermic process and the ...
2
This problem can be solved iteratively, if you are looking to increase precision. A first iteration ("zeroth order solution") was presented in another answer, where it was first assumed that approximately all reagent is converted to product before using the result to back-compute the residual (very small) concentration of reagent at equilibrium.
In ...
0
No.
For this problem I'm going to assume ideal gas conditions, mostly because looking into nonideal behavior of PCl3 and PCl5 at unspecified pressure and temperature sounds painful. There are some intermolecular forces with phosphorus compounds to consider, but I didn't see much about PCl3/PCl5 dimerization in a quick search and this table gives an entry ...
0
There is a fallacy in the discussion that in this context the notion of an equilibrium constant retains its usual meaning (as rightfully questioned in a comment). It doesn't. An equilibrium constant arises when you assume that various chemical components (in the same or different phases) can coexist. The condition for coexistence of two phases is that the ...
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