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The state in which both reactants and products are present at concentrations which have no further tendency to change with time.

9 votes
3 answers
855 views

Why is standard state useful?

We fill the container with $\ce{A}$ and after some time has passed equilibrium is established. … At least for gases it doesn't make sense because in equilibrium is their partial pressure that matters. …
Anton's user avatar
  • 766
5 votes
0 answers
38 views

Do the differential expressions of second law of thermodynamics imply a smooth increase in e... [closed]

Suppose that the system is at equilibrium state (state $1$) at time $t$. Something happens (e.g. a chemical reaction takes place) so the system will reach a new equilibrium state (state $2$). … To be honest I haven't seen any reaction reaching equilibrium in that way but I can't find a reason why it should not behave this way. …
Anton's user avatar
  • 766
2 votes
1 answer
210 views

Can we have an equilibrium constant for a reaction at constant volume?

Can the equilibrium state now be predicted? … I mean if the piston was movable (e.g. coupled to an external pressure of $1$ atm) the system would reach equilibrium and we could find the equilibrium state (e.g. the gas composition) by using the equilibrium
Anton's user avatar
  • 766
1 vote
2 answers
1k views

Is it valid to calculate concentrations given equilibrium constant and initial concentration?

Now if we are given that initial we have $x$ mol of $\ce{A}$ and no amount of $\ce{B}$ and $\ce{C}$, one can calculate the amount of B and C present at equilibrium using the equilibrium constant $K$: $ … It is a common exercise in many general Chemistry textbooks where you are given a reaction with an equilibrium constant and you must find the concentrations at equilibrium. …
Anton's user avatar
  • 766
1 vote
0 answers
182 views

Understanding Pourbaix diagram

I was taught that the lines in the Pourbaix diagrams are equilibrium lines. … The reaction describing this equilibrium line is (with equilibrium constant $K$): $$\ce{2H+_{(aq)} + 2e- <=> H2_{(g)}}$$ Now at equilibrium taking the $\log K$ and equilibrium constant we obtaing: $$\log …
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  • 766