Linked Questions
14 questions linked to/from Which equilibrium constant is appropriate to use?
3
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Thermodynamics and equilibrium constant [duplicate]
I just wanted to make sure how we can know whether it is $K_c$ (equilibrium constant of concentrations) or $K_p$ (equilibrium constant of pressures) which "comes out" of the equation, $\Delta G^o= -RT ...
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How switching values of $R$, $K_p$ and $K_c$ does not alter the value of change in standard Gibbs energy? [duplicate]
$\Delta_\text r G=RT\ln \frac{Q}{K}$
If we switch $K_p$ from $K_c$, accordingly $Q_p$ will change to $Q_c$. Therefore, value of $\Delta_\text r G$ remains the same.
$ΔG^\circ=−RT\ln K=ΔH^\circ−TΔS^\...
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How to calculate the Gibbs energy for the vaporisation of solid ammonia? [duplicate]
Given that $P_{\ce{NH3}}=\mathrm{10~atm}$ at $298\ \mathrm K$, calculate the $\Delta G^\circ$ of the reaction $$\ce{NH3(s) <=> NH3(g)}$$
I'll begin solving the problem...
I'm using the formula ...
- 33
2
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0
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Why does the pressure equilibrium constant exist? [duplicate]
Since $K_c$ can already be used for calculating the equilibrium constant for a reaction involving gases, but $K_p$ cannot be used for calculating the equilibrium constant for a reaction involving ...
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51
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3
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Unit of the equilibrium constant: contradiction of Bridgman's theorem?
The following equation is standard in thermodynamics:
$$
\Delta G^\circ=-RT\log(K)
$$
where $K$ is the equilibrium constant. In dimensional analysis, Bridgman's theorem tells us that the argument ...
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6
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5
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Why do the units of rate constants change, and what does that physically mean?
Just to be clear, I do understand that the units of the rate constant k is selected to make the equation dimensionally consistent. That is not what I am asking.
It should be the case that k 'means' ...
- 69
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What are the units of Kc and Kp?
They are both equilibrium constants as far as I know.
Kc is in terms of molarity and Kp is in terms of pressure. Also both of them are ratios of respective quantities [ ratio of molarity(s) in Kc and ...
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6
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Why are Kc and Kp interchangeable in the free energy-equilibrium equation? [duplicate]
In this equation that relates Standard free energy and the equilibrium constant:
$\Delta Gº=-RT \ln K_{eq}$
My textbook (and my teacher) say both $K_p$ (constant related to pressure) and $K_c$ (...
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3
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1
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For homogeneous equilibrium, why are liquids and solids included in the equilibrium constant (when they aren't in heterogeneous equilibria)?
In a heterogeneous reaction (where the states are varied) we do not include liquids and solids in the equilibrium equation because their concentrations do not change.
E.g. Chemguide.co.uk
However, ...
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Equilibrium constant for heterogeneous equilibria having aqueous as well as gaseous reactants
Suppose we have a heterogenous equilibrium :
$$\ce{A(aq) +B(aq) <=> C(g) +D(aq)}$$
Which equilibrium constant is used here?
Both pressure and concentration terms are there. So, according to me, ...
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3
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1
answer
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K in Gibbs free energy
When we write for an equilibrium that $$\Delta G^\circ = -RT \ln(K)$$ Which type of $K$ , ie is it $K_\text{concentration}, K_\text{pressure}$ or $K_{\chi}$ because all these have different numeric ...
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Dissociation percentages of N2O4 at different pressures using Gibbs' free energy
\begin{align}
\ce{N2O4(g)&<=>2NO2(g)} &
\Delta_\mathrm{r}G^\circ &= \pu{4.7 kJ mol^-1}
\end{align}
Knowing that the standard values for pressure and temperature are $\pu{1 bar}$...
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Equilibrium constants and
I have a few questions about this two relations:
We know that:
$K_p$ = $e^{\frac{-\Delta G^0}{RT}}$ $(1)$
and
$K_c = K_p (RT)^{\Delta n}$ $(2)$
Ok now I'm solving this problem about solubility:
The ...
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Molarity of solids and liquids in equilibrium constant expression [duplicate]
Why is the molarity of pure liquids , solids and excess solvent taken 1 in the equilibrium constant expression?
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