Equilibrium Concentration


The algebraic form of the equilibrium expression (and reaction quotient)

\[ \dfrac{[C]^c[D]^d}{[A]^a[B]^b} = Q~\mathrm{or}~K \]

allows us to solve for various terms in the equation given other known quantities. For example, if the following reaction

\[ \mathrm{N_2O_4}(g) \rightleftharpoons 2\mathrm{NO}_2 (g) \]

had, at equilibrium, the following equilibrium concentrations \[\begin{align*} [\mathrm{N_2O_4}]_{\mathrm{eq}} &= 0.670~M \\ [\mathrm{NO_2}]_{\mathrm{eq}} &= 0.0547~M \end{align*}\] one could determine the equilibrium constant, K, algebraically as follows

\[\begin{align*} K &= \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} \\[1ex] &= \dfrac{(0.0547)^2}{0.670} \\[1ex] &= 4.46\times 10^{-3} \end{align*}\]

However, if we knew K and one of the equilibrium concentrations, we could solve for the missing concentration.

Often times we only know the value of K and the initial concentrations of the species in the reaction. Although it may not be immediately noticeable, we can still solve for the equilibrium concentrations in this case. The reason for this is due to the stoichiometric relationships of the reactant and products.


Solving Equilibrium Problems


Let us consider again, the reaction above with the following experimental data

\[\begin{align*} \mathrm{N_2O_4}(g) &\rightleftharpoons 2\mathrm{NO}_2 (g) \\[2ex] \mathrm{[N_2O_4]_i} &= 0.100 \\[1ex] \mathrm{[NO_2]_i} &= 0.000 \\[1ex] K &= 4.46\times 10^{-3} \end{align*}\]

Here we start the reaction with only [N2O4]eq and no product. How do we get the equilibrium concentrations from this?


ICE Tables

An ICE (or RICE) table has the following rows

  • R = reaction
  • I = initial concentration (M)
  • C = change in concentration (M)
  • E = equilibrium concentration (M)

It is important to note that an ICE table is not some special, magical thing that you must do to solve these types of problems. The only point of an ICE table is to keep your information organized. That is it.


Fill in givens

We will now write the ICE table for our reaction and include the given information.

N2O4 \(\rightleftharpoons\) 2NO2
I 0.100 0
C
E


Compare Q to K

To determine C (i.e. the change in concentration row), you must compare Q and K and decide which way the reaction will go (left or right). This represents how much the reactant and product concentrations will change to reach the equilibrium concentrations. Since we are only starting with reactant and no product, no math is required as this reaction can only go toward the right. We will represent our change in concentration using x and stoichiometry.

N2O4 \(\rightleftharpoons\) 2NO2
I 0.100 0
C -x +2x
E

Notice that since the reaction is proceeding to the right, we will experience a decrease in our reactant concentration (hence the “-x”). Similarly, product will be produced but two times more than the reactant consumption due to stoichiometry. For every one mole of rectant consumed, two moles of product are produced. Therefore, I write in a “+2x”.


Fill in the last row

Finally, to get the last row (E), simply combine the first two columns.

N2O4 \(\rightleftharpoons\) 2NO2
I 0.100 0
C -x +2x
E 0.100 – x 2x


Write the equilibrium expression

Write out the equilibrium expression.

\[\dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} = K\]

Substitute in the equilibrium concentration expressions you wrote in row E of the ICE table as well as K.

\[\dfrac{(2x )^2}{(0.100 - x)} = 4.46\times 10^{-3}\]


Solve for x

There are a variety of ways to solve for x depending on the form of the equlibrium expression. Some methods of solving for x are straight forward algebraic exercises while some approaches are faster but do not always work. We will walk through the different ways of solving for x here using our equilibrium expression above.


Small x approximation

To avoid the quadratic formula, you can try to use the small x apporoximation by applying some chemical intuition. Recall our discussion regarding the magnitude of the equliibrium constant, K. For our reaction, K is rather small (much less than one), therefore the reaction is “reactant favored”. That is, at equilibrium, we would expect to have much more reactants than products.

Now ask yourself, are we starting this reaction near equilibrium? If, at equilibrium, we should have an abundance of reactant, and we are starting our reaction with all reactant, then we should be starting close to equliibrium! In this case, x should be small. The question is, is x small enough?

Consider our equilibrium expression once again

\[\dfrac{(2x )^2}{(0.100 - x)} = 4.46\times 10^{-3}\]

If x is small, then we could tweak the equation to give

\[\dfrac{(2x )^2}{0.100} = 4.46\times 10^{-3}\]

What we have done is made the assumption that x is so small that \((0.100-x) \approx 0.100\). We will see if we assumed correctly later.

Next, solve directly for x.

\[\begin{align*} \dfrac{(2x )^2}{(0.100)} &= 4.46\times 10^{-3} \\[1ex] 4x^2 &= 4.46\times 10^{-4}\\[1ex] x^2 &= 1.115\times 10^{-4} \\[1ex] x &= \sqrt{1.115\times 10^{-4}} \\[1ex] &= 1.06\times 10^{-2} \end{align*}\]

TEST x

At this point we STOP and text x. The rule I use is, if x is less than 5% of the initial concentration we were subtracting it from (or adding it to), then x is small enough to justify using the small x approximation. Let us test it now.

\[\begin{align*} \dfrac{x}{0.100} \times 100\% \longrightarrow \\[1ex] \dfrac{1.056\times 10^{-2}}{0.100} \times 100 = 10.6\% \end{align*}\]

Here, x was too big (greater than 5%) so therefore we should not use the x we obtained using the small x approximation. Solve for x using the quadratic formula way.


Quadratic formula

Note here that we will have the form of a quadratic equation (easy to spot since we will have an x2 term and an x term). So let us rearrange this equation into quadratic form.

\[\begin{align*}\require{color} 4x^2 &= 4.46\times 10^{-3}(0.100 -x) \\[1ex] &= 4.46\times 10^{-4} - 4.46\times 10^{-3}x \\[1ex] \textcolor{red}{4}x^2 + \textcolor{blue}{4.46\times 10^{-3}}x \textcolor{green}{-4.46\times 10^{-4}} &= 0 \end{align*}\]

I have highlighed the coefficients to denote \[\begin{align*} a &= \textcolor{red}{4}\\ b &= \textcolor{blue}{4.46\times 10^{-3}}\\ c &= \textcolor{green}{-4.46\times 10^{-4}} \end{align*}\]

for the quadratic formula

\[x = \dfrac{-\textcolor{blue}{b}\pm\sqrt{\textcolor{blue}{b}^2-4\textcolor{red}{a}\textcolor{green}{c}}}{2\textcolor{red}{a}}\]

Plug in the values for each coefficient and solve for x.

\[\begin{align*} x &= \dfrac{-\textcolor{blue}{b}\pm\sqrt{\textcolor{blue}{b}^2-4\textcolor{red}{a}\textcolor{green}{c}}}{2\textcolor{red}{a}}\\[2ex] &= \dfrac{-\textcolor{blue}{4.46\times 10^{-3}} \pm\sqrt{(\textcolor{blue}{4.46\times 10^{-3}})^2 - 4(\textcolor{red}{4})(\textcolor{green}{-4.46\times 10^{-4}})}}{2(\textcolor{red}{4})}\\[2ex] &\approx 0.01002~~\mathrm{or}~~-0.01113 \end{align*}\]

Ignore the negative root and take the postive root for x. Therefore,

\[ x = 0.01002 \]


Solve for the equilibrium concentrations

Return to your ICE table

N2O4 \(\rightleftharpoons\) 2NO2
I 0.100 0.000
C -x +2x
E 0.100 - x 2x

and plug in your value for x that was obtained using the quadratic formula since the small x approximation did not work.

N2O4 \(\rightleftharpoons\) 2NO2
I 0.100 0.000
C -x +2x
E 0.100 - 0.01002 2(0.01002)

Solve for the equilibrium concentrations

\[\begin{align*} [\mathrm{N_2O_4}]_{\mathrm{eq}} &= 0.100 - 0.01002 = 0.0900~M \\[1ex] [\mathrm{NO_2}]_{\mathrm{eq}} &= 2(0.01002) = 0.0200~M \end{align*}\]

EXERCISE: Go back and recompute the equilibrium concentrations with the x we got from using the small x approximation. We said that our x was too large and should not be used. How far off would these incorrect equilibrium concentrations be?


Practice Problems


Problem 1

Given is a reaction as well as a table of reaction data for five different experiments. Initial reactant and product concentrations and equilibrium concentrations are given as well as the equilibrium constants.

\[\begin{align*} \mathrm{N_2O_4}(g) \rightleftharpoons 2\mathrm{NO}_2 (g) \end{align*}\]

Initial and equilibrium concentrations (in M) of N2O4 and NO2 at 25 °C
Experiment [N2O4]i [NO2]i [N2O4]eq [NO2]eq K
1 0.670 0.000 0.643 0.0547 \(4.65\times 10^{-3}\)
2 0.446 0.0500 0.448 0.0457 \(4.66\times 10^{-3}\)
3 0.500 0.0300 0.491 0.0475 \(4.60\times 10^{-3}\)
4 0.600 0.0400 0.594 0.0523 \(4.60\times 10^{-3}\)
5 0.000 0.200 0.0898 0.0204 \(4.63\times 10^{-3}\)

Solve for the equilibrium concentrations for each experiment (given in columns 4 and 5). Use the small x approximation where appropriate; otherwise use the quadratic formula. Refer to the equilibrium concentrations in the table to check your answers.


Solution Experiment 1

No product. Shift right. \[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= K \\[1ex] \dfrac{(2x)^2}{0.670-x} &= 4.65\times 10^{-3} \end{align*}\] Quadratic approach: \[\begin{align*} 4x^2 &= 4.65\times 10^{-3} (0.670-x) \\ 4x^2 &= 3.12\times 10^{-3} - 4.65\times 10^{-3}x \\ 4x^2 + 4.65\times 10^{-3}x - 3.12\times 10^{-3} &= 0 \\[2ex] x &= \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\ &= \dfrac{-4.65\times 10^{-3} \pm \sqrt{(4.65\times 10^{-3})^2 - 4(4)(- 3.12\times 10^{-3})}}{2(4)} \\ &= 0.0274 \quad \mathrm{and} \quad -0.0285 \end{align*}\] First root: \[\begin{align*} [\mathrm{N_2O_4}] &= 0.670 - x = 0.670 - 0.0274 = 0.643 \\ [\mathrm{NO_2}] &= 2x = 2(0.0274) = 0.0548 \end{align*}\]

Small \(x\) approach (less than 5%): \[\begin{align*} \dfrac{(2x)^2}{0.670} &= 4.65\times 10^{-3} \\ 4x^2 &= (4.65\times 10^{-3})(0.670) \\ x &= \sqrt{\dfrac{(4.65\times 10^{-3})(0.670)}{4}} \\ &= 0.0279 \\[2ex] \dfrac{x}{[\mathrm{N_2O_4}]_{\mathrm{i}}} \times 100\% &= \dfrac{0.0279}{0.670}\times 100\% = 4.16\%\\[2ex] [\mathrm{N_2O_4}] &= 0.670 - x = 0.670 - 0.0279 = 0.642 \\ [\mathrm{NO_2}] &= 2x = 2(0.0279) = 0.0558 \end{align*}\]


Solution Experiment 2

\[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= Q \\[1ex] \dfrac{(0.0500)^2}{0.446} &= \\ 5.61\times 10^{-3} &= \end{align*}\] \(Q > K\); \(5.61\times 10^{-3} > 4.66\times 10^{-3}\) – Shift left \[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= K \\[1ex] \dfrac{(0.0500-2x)^2}{0.446+x} &= 4.66\times 10^{-3} \end{align*}\] Quadratic approach: \[\begin{align*} 2.5\times 10^{-3} - 0.1x - 0.1x + 4x^2 &= 4.66\times 10^{-3} (0.446 + x) \\ 2.5\times 10^{-3} - 0.2x + 4x^2 &= 2.08\times 10^{-3} + 4.66\times 10^{-3}x \\ 4x^2 - 0.205x + 4.2\times 10^{-4} &= 0 \end{align*}\] \[\begin{align*} x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} &= \dfrac{-(-0.205) \pm \sqrt{(-0.205)^2 - 4(4)(4.2\times 10^{-4})}}{2(4)} \\ &= \dfrac{0.205 \pm \sqrt{3.53\times 10^{-2}}}{8} \\ &= 4.91 \times 10^{-2} \quad \mathrm{and} \quad 2.14\times 10^{-3} \end{align*}\] First root: \[\begin{align*} [\mathrm{N_2O_4}] &= 0.446 + x = 0.446 + 4.91\times 10^{-2} = 0.495 \\ [\mathrm{NO_2}] &= 0.0500-2x = 0.0500 - 2(4.91\times 10^{-2}) = -0.0482 \quad \mathbf{UNPHYSICAL} \end{align*}\] Second root: \[\begin{align*} [\mathrm{N_2O_4}] &= 0.446 + x = 0.446 + 2.14\times 10^{-3} = 0.448 \\ [\mathrm{NO_2}] &= 0.0500-2x = 0.0500 - 2(2.14\times 10^{-3}) = 0.0457\\[2ex] \dfrac{(0.0457)^2}{0.448} &= 4.66\times 10^{-3} \quad \mathbf{RIGHT} \end{align*}\]

Small \(x\) approach (less than 5%): \[\begin{align*} \dfrac{(0.0500-2x)^2}{0.446} &= 4.66\times 10^{-3} \\ (0.0500-2x)^2 &= 2.08\times 10^{-3} \\ 0.0500-2x &= 4.56\times 10^{-2} \\ -2x &= -4.41\times 10^{-3} \\ x &= 2.205\times 10^{-3} \\ \dfrac{x}{[\mathrm{NO_2}]_{\mathrm{i}}} \times 100\% &= \dfrac{2.205\times 10^{-3}}{0.446}\times 100\% = 0.49\%\\[2ex] [\mathrm{N_2O_4}] &= 0.446 + x = 0.446 + 2.205\times 10^{-3} = 0.448 \\ [\mathrm{NO_2}] &= 0.0500-2x = 0.0500 - 2(2.205\times 10^{-3}) = 0.046 \end{align*}\]


Solution Experiment 3

\[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= Q \\[1ex] \dfrac{(0.0300)^2}{0.500} &= \\ 1.8\times 10^{-3} &= \end{align*}\] \(Q < K\); \(1.81\times 10^{-3} < 4.60\times 10^{-3}\) – Shift right \[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= K \\[1ex] \dfrac{(0.0300+2x)^2}{0.500-x} &= 4.60\times 10^{-3} \end{align*}\] Quadratic approach: \[\begin{align*} 9.0\times 10^{-4} + 0.06x + 0.06x + 4x^2 &= 4.60\times 10^{-3} (0.500 - x) \\ 9.0\times 10^{-4} + 0.12x + 4x^2 &= 2.3\times 10^{-3} - 4.60\times 10^{-3}x \\ 4x^2 + 0.1246x - 1.4\times 10^{-3} &= 0 \end{align*}\] \[\begin{align*} x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} &= \dfrac{-0.1246 \pm \sqrt{(0.1246)^2 - 4(4)(-1.4\times 10^{-3})}}{2(4)} \\ &= \dfrac{-0.1246 \pm \sqrt{3.79\times 10^{-2}}}{8} \\ &= 8.77 \times 10^{-3} \quad \mathrm{and} \quad -3.99\times 10^{-2} \end{align*}\] First root: \[\begin{align*} [\mathrm{N_2O_4}] &= 0.500 - x = 0.500 - 8.77\times 10^{-3} = 0.491 \\ [\mathrm{NO_2}] &= 0.0300+2x = 0.0300 + 2(8.77\times 10^{-3}) = 0.0475 \end{align*}\] Small \(x\) approach (less than 5%): \[\begin{align*} \dfrac{(0.0300-2x)^2}{0.500} &= 4.60\times 10^{-3} \\ (0.0300+2x)^2 &= 2.3\times 10^{-3} \\ 0.0300+2x &= 4.80\times 10^{-2} \\ 2x &= 1.80\times 10^{-2} \\ x &= 8.98\times 10^{-3} \\ \dfrac{x}{[\mathrm{NO_2}]_{\mathrm{i}}} \times 100\% &= \dfrac{8.98\times 10^{-3}}{0.500}\times 100\% = 1.8\% \\[2ex] [\mathrm{N_2O_4}] &= 0.500 - x = 0.500 - 8.89\times 10^{-3} = 0.491 \\ [\mathrm{NO_2}] &= 0.0300+2x = 0.0300 + 2(8.89\times 10^{-3}) = 0.0478 \end{align*}\]


Solution Experiment 4

\[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= Q \\[1ex] \dfrac{(0.0400)^2}{0.600} &= \\ 2.67\times 10^{-3} &= \end{align*}\] \(Q < K\); \(2.67\times 10^{-3} < 4.60\times 10^{-3}\) – Shift right \[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= K \\[1ex] \dfrac{(0.0400+2x)^2}{0.600-x} &= 4.60\times 10^{-3} \end{align*}\] Quadratic approach: \[\begin{align*} 1.6\times 10^{-3} + 0.08x + 0.08x + 4x^2 &= 4.60\times 10^{-3} (0.600 - x) \\ 9.0\times 10^{-4} + 0.16x + 4x^2 &= 2.76\times 10^{-3} - 4.60\times 10^{-3}x \\ 4x^2 + 0.1646x - 1.16\times 10^{-3} &= 0 \end{align*}\] \[\begin{align*} x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} &= \dfrac{-0.1646 \pm \sqrt{(0.1646)^2 - 4(4)(-1.16\times 10^{-3})}}{2(4)} \\ &= \dfrac{-0.1646 \pm \sqrt{4.565\times 10^{-2}}}{8} \\ &= 6.13 \times 10^{-3} \quad \mathrm{and} \quad -4.73\times 10^{-2} \end{align*}\] First root: \[\begin{align*} [\mathrm{N_2O_4}] &= 0.600 - x = 0.600 - 6.13\times 10^{-3} = 0.594 \\ [\mathrm{NO_2}] &= 0.0400+2x = 0.0400 + 2(6.13\times 10^{-3}) = 0.0523 \end{align*}\] Small \(x\) approach (less than 5%): \[\begin{align*} \dfrac{(0.0400-2x)^2}{0.600} &= 4.60\times 10^{-3} \\ (0.0400+2x)^2 &= 2.76\times 10^{-3} \\ 0.0400+2x &= 5.25\times 10^{-2} \\ 2x &= 1.25\times 10^{-2} \\ x &= 6.27\times 10^{-3} \\ \dfrac{x}{[\mathrm{NO_2}]_{\mathrm{i}}} \times 100\% &= \dfrac{6.27\times 10^{-3}}{0.600}\times 100\% = 1.12\%\\[2ex] [\mathrm{N_2O_4}] &= 0.600 - x = 0.600 - 6.27\times 10^{-3} = 0.594 \\ [\mathrm{NO_2}] &= 0.0400+2x = 0.0400 + 2(6.27\times 10^{-3}) = 0.0525 \end{align*}\]


Solution Experiment 5

No reactant. Shift left. \[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= K \\[1ex] \dfrac{(0.200-2x)^2}{x} &= 4.63\times 10^{-3} \end{align*}\] Quadratic approach: \[\begin{align*} 4.0\times 10^{-2} - 0.4x - 0.4x + 4x^2 &= 4.63\times 10^{-3}x \\ 4.0\times 10^{-2} - 0.8x + 4x^2 &= 4.63\times 10^{-3}x \\ 4x^2 - 0.8046x + 4.0\times 10^{-2} &= 0 \end{align*}\] \[\begin{align*} x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} &= \dfrac{-(-0.8046) \pm \sqrt{(-0.8046)^2 - 4(4)(4.0\times 10^{-2})}}{2(4)} \\ &= \dfrac{0.8046 \pm \sqrt{7.381\times 10^{-3}}}{8} \\ &= 1.11 \times 10^{-1} \quad \mathrm{and} \quad 8.98\times 10^{-2} \end{align*}\] First root: \[\begin{align*} [\mathrm{N_2O_4}] &= x = 0.111 \\ [\mathrm{NO_2}] &= 0.200-2x = 0.200 - 2(1.11\times 10^{-1}) = -0.022 \quad \mathbf{UNPHYSICAL} \end{align*}\] Second root: \[\begin{align*} [\mathrm{N_2O_4}] &= x = 0.0898 \\ [\mathrm{NO_2}] &= 0.200-2x = 0.200 - 2(8.98\times 10^{-2}) = 0.0204 \end{align*}\] Small \(x\) approach (less than 5%): \[\begin{align*} \dfrac{(0.200-2x)^2}{x} &= 4.63\times 10^{-3} \\ \dfrac{(0.200)^2}{x} &= 4.63\times 10^{-3} \\ x &= 8.64 \\ \dfrac{x}{[\mathrm{NO_2}]_{\mathrm{i}}} \times 100\% &= \dfrac{8.64}{0.200}\times 100\% = 4.32\times 10^3\% \quad \mathbf{TOO~LARGE} \end{align*}\]


Problem 2

Find the equilibrium concentration for the following reaction.

\[\begin{align*} \mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons \mathrm{2HI}(g) \end{align*}\] \[\begin{align*} [\mathrm{H_2}]_{\mathrm{i}} = 0.240~M \quad [\mathrm{I_2}]_{\mathrm{i}} = 0.240~M \quad K = 54.3 \end{align*}\]

Solution

Make ICE table. Proceeds right.

H2 + I2 \(\rightleftharpoons\) 2HI
I 0.240 0.240 0.000
C -x -x +2x
E 0.240 – x 0.240 – x 2x


\[\begin{align*} \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} &= K \\ \dfrac{(2x)^2}{(0.240-x)(0.240-x)} &= 54.3 \\ \dfrac{(2x)^2}{(0.240-x)^2} &= 54.3 \\ \dfrac{2x}{0.240-x} &= 7.369 \\ 2x &= 7.369 (0.240 - x) \\ &= 1.768 - 7.369x \\ 9.369x &= 1.768 \\ x &= 0.1887 \\[2ex] [\mathrm{H_2}]_{\mathrm{eq}} &= 0.240 - x = 0.240 - 0.1887 = 0.0513~M \\ [\mathrm{I_2}]_{\mathrm{eq}} &= 0.240 - x = 0.240 - 0.1887 = 0.0513~M \\ [\mathrm{HI}]_{\mathrm{eq}} &= 2x = 2(0.1887) = 0.377~M \end{align*}\]


Problem 3

Find the equilibrium concentration for the following reaction.

\[\begin{align*} \mathrm{N_2}(g) + \mathrm{O_2}(g) \rightleftharpoons \mathrm{2NO}(g) \end{align*}\] \[\begin{align*} [\mathrm{N_2}]_{\mathrm{eq}} = 0.036~M \quad [\mathrm{O_2}]_{\mathrm{eq}} = 0.0089~M \quad [\mathrm{NO}]_{\mathrm{eq}} = ~? \quad K = 4.1\times 10^{-4} \end{align*}\]

Solution

No need for ICE table. No initial concentrations. Just solve for \([\mathrm{NO}]\) in equilibrium expression. \[\begin{align*} \dfrac{[\mathrm{NO}]^2}{[\mathrm{N_2}][\mathrm{O_2}]} &= K \\ \dfrac{[\mathrm{NO}]^2}{(0.036)(0.0089)} &= 4.1\times 10^{-4} \\ [\mathrm{NO}] &= 3.6\times 10^{-4} \end{align*}\]


Problem 4

Find the equilibrium concentration for the following reaction.

\[\begin{align*} \mathrm{PCl_5}(g) \rightleftharpoons \mathrm{PCl_3}(g) + \mathrm{Cl_2}(g) \end{align*}\] \[\begin{align*} [\mathrm{PCl_5}]_{\mathrm{i}} = 1.00 \quad [\mathrm{PCl_3}]_{\mathrm{eq}} = ~? \quad [\mathrm{Cl_2}]_{\mathrm{eq}} = ~? \quad K = 0.00211 \end{align*}\]

Solution

Make ICE table. Reaction proceeds right.

PCl5 \(\rightleftharpoons\) PCl3 + Cl2
I 1.00 0 0
C -x +x + x
E 1.00 – x x x


\[\begin{align*} \dfrac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]} &= K \\ \dfrac{(x)(x)}{1.00-x} &= 0.00211 \end{align*}\] Small \(x\) approximation: \[\begin{align*} \dfrac{x^2}{1.00} &= 0.00211 \\ x^2 &= 0.00211 \\ x &= 0.0459 \end{align*}\] Test \(x\): \[\begin{align*} \dfrac{x}{[\mathrm{PCl_5}]_{\mathrm{i}}} &= \dfrac{0.0459}{1.00}\times 100\% = 4.59\% \end{align*}\] Find equilibrium concentrations: \[\begin{align*} [\mathrm{PCl_5}]_{\mathrm{eq}} &= 1.00 - x = 1.00 - 0.0459 = 0.954\\ [\mathrm{PCl_3}]_{\mathrm{eq}} &= x = 0.0459 \\ [\mathrm{Cl_2}]_{\mathrm{eq}} &= x = 0.0459 \end{align*}\]


Problem 5

Find the equilibrium constant, K, for the following reaction.

\[\begin{align*} \mathrm{I_2}(aq) + \mathrm{I^-}(aq) \rightleftharpoons \mathrm{I_3^-}(aq) \end{align*}\] \[\begin{align*} [\mathrm{I_2}]_{\mathrm{i}} &= 1.00\times 10^{-3}~M \\ [\mathrm{I^-}]_{\mathrm{i}} &= 1.00\times 10^{-3}~M \\ [\mathrm{I_2}]_{\mathrm{eq}} &= 6.61\times 10^{-4}~M \\[2ex] K &= ~? \end{align*}\]

Solution

Make ICE table. Realize that \(x\) can be found directly by taking [I\(_2\)]\(_{\mathrm{i}}\) \(-\) [I\(_2\)]\(_{\mathrm{eq}}\).

I2 + I \(\rightleftharpoons\) I3
I 0.001 0.001 0
C -x -x + x
E 0.000661 0.001–x x


\[\begin{align*} [\mathrm{I_2}]_{\mathrm{i}} - x &= [\mathrm{I_2}]_{\mathrm{eq}} \\ x &= [\mathrm{I_2}]_{\mathrm{i}} - [\mathrm{I_2}]_{\mathrm{eq}} \\ &= 1.00\times 10^{-3} - 6.61\times 10^{-4} \\ &= 3.39\times 10^{-4}~M\\[2ex] K &= \dfrac{[\mathrm{I_3^-}]}{[\mathrm{I_2}][\mathrm{I^-}]} \\ &= \dfrac{x}{(6.61\times 10^{-4})(1.00\times 10^{-3}-x)} \\ &= \dfrac{3.39\times 10^{-4}}{(6.61\times 10^{-4})(1.00\times 10^{-3}-3.39\times 10^{-4})} \\ &= 776 \end{align*}\]


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