All Questions
18
questions
0
votes
1
answer
906
views
Standard Gibbs free energy when all the reactants and products are at standard condition
I have read in my textbook that for a reaction $\Delta G=\Delta G^o-RT\ln Q $, where $\Delta G^o$ is the Gibbs free energy change when the initial concentration of products and reactants are unity. ...
- 159
1
vote
4
answers
3k
views
Is Gibbs free energy change applicable to forward and reverse reactions at equilibrium?
Consider a reaction
$$\ce{A + B <=> C + D}. \tag{R1}$$
Now $\Delta G$ for the forward reaction is
$$\Delta G_\mathrm{fwd} = \Delta H - T\Delta S. \tag{1}$$
For the reverse reaction $\Delta H$ ...
- 19
0
votes
0
answers
245
views
Gibbs free energy of phosphorus pentachloride decomposition reaction
The equilibrium constant at $\pu{227 °C}$ for the equation
$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$
is $K_p = \pu{4.50E3 bar}.$ Calculate the value of $Δ_\mathrm{rxn}G^\circ$ at $\pu{227 ...
- 39
1
vote
0
answers
57
views
Calculating thermodynamic quantities for hydrogenation of iron nitride
HW#6.3 If the reaction
$$\ce{Fe2N(s) + 3/2 H2(g) <=> 2 Fe(s) + NH3(g)}$$
comes to an equilibrium at a total pressure of $\pu{1 bar},$ analysis of the gas shows that at $\pu{700 K}$ and $\pu{800 ...
2
votes
3
answers
1k
views
How to explain disagreement between Le Châtelier's principle and the simplified Gibbs free energy equation?
For example, for the dissolution of a salt in water that is exothermic, heating the solution would drive the reaction towards the solid form of the salt according to Le Châtelier's principle.
However,...
- 101
2
votes
2
answers
3k
views
Gibbs free energy in standard state vs. equilibrium
I have a problem with the definition of the standard Gibbs energy and its connection to the equilibrium constants.
I think, that I've basically understood what the different equation mean but there ...
- 63
2
votes
0
answers
584
views
Conversion of graphite into diamond [duplicate]
The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at $T = \pu{298 K}$ are $\pu{0 kJ mol-1}$ and $\pu{2.9 kJ mol-1}$, respectively.
The conversion of graphite [C(...
- 67
8
votes
2
answers
1k
views
Why proton concentration is divided by 10⁻⁷?
I am reviewing the book Biochemistry Concepts and Connections by Appling, Cahill, and Mathews and I cannot understand why they divide by the hydrogen concentration by $10^{-7}$. Why not just leave it ...
- 191
6
votes
1
answer
6k
views
Pressure at which graphite and diamond are in equilibrium
The standard state Gibbs free energies of formation of graphite and diamond at $T = \pu{298 K}$ are $\pu{0 kJ mol-1}$ and $\pu{2.9 kJ mol-1}$, respectively.
The conversion of graphite to diamond ...
- 407
2
votes
1
answer
143
views
How to calculate the standard Gibbs energy at equilibrium?
Given that $K_c = 1.7 \times 10^{-13}$, calculate $\Delta G^{\circ}$ for this equilibrium mixture at $\pu{298 K}$.
$$\ce{N2O (g) + \frac{1}{2}O2 (g) <=> 2 NO (g)}$$
I've calculated:
$$
\begin{...
- 131
3
votes
2
answers
20k
views
How to calculate Gibbs free energy from pKa?
Is it possible to calculate Gibbs free energy ($ΔG = -RT\ln K_\mathrm{eq}$) of an acid dissociation reaction using the $\mathrm{p}K_\mathrm{a}$? If so, how?
- 41
2
votes
0
answers
540
views
Active mass,Pressure and concentration
The concentration is connected to the activity via
$$a(\ce{A})= \gamma_{c,\ce{A}}\cdot{}\frac{c(\ce{A})}{c^\circ},$$
where the standard concentration is $c^\circ = 1\:\mathrm{mol/L}$. At reasonable ...
4
votes
2
answers
505
views
Equilibrium constant. Can it be reached?
Consider the reaction below:
$$\ce{A + B <=> C}$$
Suppose that the equilibrium constant for this reaction is $K = 10$.
I then prepare a reaction vessel with volume of $\pu{1 dm^{-3}}$ which ...
- 41
25
votes
2
answers
3k
views
Which equilibrium constant is appropriate to use?
I have learnt that the standard free energy change is related to the equilibrium constant of a reaction by,
$$\Delta G^\circ = -RT \ln K$$
Here, does $K$ refer to $K_p$ or $K_c$?
Also, please give ...
- 1,245
14
votes
4
answers
4k
views
Is there a reason for the mathematical form of the equilibrium constant? [duplicate]
Why are the two molarities multiplied and not added, and why is each raised to the power of the coefficient rather than multiplied by it? What is the reasoning behind this form? Was it simply ...
- 939