Linked Questions

3 votes
1 answer
2k views

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 ...
Pao's user avatar
  • 243
0 votes
1 answer
79 views

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^\...
Apurvium's user avatar
  • 1,270
1 vote
1 answer
188 views

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 ...
user259462's user avatar
2 votes
0 answers
57 views

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 ...
Frank's user avatar
  • 380
52 votes
3 answers
5k views

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 ...
John Gowers's user avatar
6 votes
5 answers
2k views

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' ...
Aerovolo's user avatar
0 votes
3 answers
25k views

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 ...
Gamira's user avatar
  • 177
6 votes
1 answer
14k views

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 $K_c$ (...
Nuria's user avatar
  • 275
3 votes
1 answer
6k views

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, ...
K-Feldspar's user avatar
  • 2,853
4 votes
1 answer
1k views

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, ...
Aditya Prakash's user avatar
3 votes
1 answer
1k views

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 ...
Mr.HiggsBoson's user avatar
3 votes
0 answers
817 views

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 $\pu{1 bar}$...
L3ul's user avatar
  • 587
0 votes
0 answers
39 views

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 ...
Granger Obliviate's user avatar
1 vote
0 answers
31 views

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?
Jasgeet Singh's user avatar