Calculating Equilibrium Concentrations
Introduced| English |
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| approximation/əˌprɒksɪˈmeɪʃn/ |
Solving for the unknown amounts
- Given the starting amounts and $K$, can you predict the end?
- Yes -- set up a table with an unknown change "$x$."
- Solve the equation $K$ gives you for $x$.
- Then read off exactly how much of each is present.
Let the change be x
- Build an ICE table with the change written as $x$.
- Reactants lose $x$; products gain $x$, each times its coefficient.
- The equilibrium row is written in terms of $x$.
In an ICE table, a reactant's equilibrium amount is written as...
Reactants are used up, so their amount is (initial $- x$).
In an ICE table, a product's equilibrium amount is written as initial plus ____.
Products are formed, so their amount is (initial $+ x$).
Solve the K equation
- Put the equilibrium row into the $K$ expression.
- Set it equal to the known $K$ and solve for $x$.
- Then plug $x$ back to get each concentration.
To find $x$, you set the $K$ expression equal to...
Set the expression equal to $K$ and solve for $x$.
A handy approximation
- When $K$ is very small, $x$ is tiny compared to the start.
- You can often use this approximation 近似 and ignore $x$ in "$(\text{start} - x)$."
- This avoids the quadratic and still gives a good answer.
Solve for equilibrium amounts
Use x to find the equilibrium concentrations from K.
When $K$ is very small, you can often ignore $x$ in (initial $- x$).
A small $K$ makes $x$ negligible compared to the start.
Start with $1.0\ \text{M}$ reactant and a small $K$. If $x$ is tiny, what is the reactant concentration?
- $(1.0 - x) \approx 1.0$ when $x$ is negligible.
- So the reactant stays close to $1.0\ \text{M}$.
The "ignore $x$" approximation should be checked by confirming $x$ is under about...
If $x$ exceeds about 5%, solve the full quadratic instead.
A solution for $x$ that gives a negative concentration should be rejected.
Concentrations cannot be negative, so that root is unphysical.
The "ignore $x$" shortcut is valid only when $K$ is small (so $x$ is much less than the starting amount) -- check that $x$ is under about 5% of the start, or solve the full quadratic. Set the ICE signs correctly: reactants minus $x$, products plus $x$. And confirm $x$ is physically reasonable (no negative concentrations).
To find equilibrium concentrations, write the change as $x$ in an ICE table, put the equilibrium row into the $K$ expression, and solve for $x$. When $K$ is small, an approximation ($x$ negligible) avoids the quadratic -- but only if $x$ stays under about 5% of the start.