Exercises
Exercise 1. The following reaction has Kc = 4.60 x 10-3 at a certain temperature.
N2O4 (g) ⇄ 2 NO2 (g)
The initial concentration of N2O4 was 3.60 M. What are the concentrations of N2O4 and NO2 at equilibrium?
Set up an ICE table.
\(\displaystyle K_c\;=\;\frac{[NO_2]^2}{[N_2O_4]}\;=\;4.60 \times 10^{-3}\)
\(\displaystyle \frac{(2x)^2}{(3.60\;-\;x)}\;=\;4.60 \times 10^{-3}\)
Notice that Kc is very small compared to a concentration of 3.60 M. We can make the assumption that x is negligible and 3.60 M – x ≅ 3.60 M. The concentration of x will not affect the initial concentration. To simplify the math, we can neglect x in the denominator.
\(\displaystyle \frac{(2x)^2}{(3.60)}\;=\;4.60 \times 10^{-3}\)
Solve the equation for x.
\(\displaystyle 4x^2\;=\;3.60 \times (4.60 \times 10^{-3})\)
\(x\;=\;\sqrt{\frac {3.60 \times (4.60 \times 10^{-3})}{4}}\;=\;0.0643\;M\)
Next we check the assumption using the 5% rule.
\(\displaystyle \frac{0.0643\;M}{3.60\;M}\times 100\;=\;1.8%\)
The assumption is valid as 1.8% < 5%.
Calculate the concentrations at equilibrium.
[N2O4]eq = 3.60 M – x = 3.60 M – 0.0643 M = 3.54 M
[NO2]eq = 2x = 2 x 0.0643 M = 0.129 M
Back to The Simplifying Assumption
Exercise 2. The following reaction has Kp = 0.0503.
2 Fe (l) + 3 CO2 (g) ⇄ Fe2O3 (s) + 3 CO (g)
Calculate the pressures of all gas species at equilibrium if the initial pressure of CO2 was 28.25 atm.
Set up an ICE table.
\(\displaystyle K_p\;=\;\frac {[CO]^3}{[CO_2]^3}\;=\;0.0503\)
Make the assumption that 28.25 atm – 3x ≅ 28.25 atm
\(\displaystyle \frac {(3x)^3}{(28.25)^3}\;=\;0.0503\)
Solve for x.
\(\displaystyle x\;=\;\Biggl(\frac{0.0503\times (2.254\times 10^4)}{27}\Biggr)^{1/3}\;=\;3.48\)
Check assumption using the 5% rule.
\(\displaystyle \frac{3.48\;atm}{28.25\;atm}\times 100\;=\;12\%\;>\;5%\) We cannot neglect x. We must solve the equation for x.
\(\displaystyle \frac {(3x)^3}{(28.25\;-\;3x)^3}\;=\;0.0503\)
We have a perfect cube on the left hand side of the equation. Take the cubed root of both sides of the equation.
\(\displaystyle \Biggl(\frac {(3x)^3}{(28.25\;-\;3x)^3}\Biggr)^{1/3}\;=\;0.0503^{1/3}\)
\(\displaystyle \frac {3x}{(28.25\;-\;3x)}\;=\;0.369\)
\(\displaystyle 3x\;=\;(0.369) \times\;(28.25\;-\;3x)\)
\(\displaystyle 3x\;=\;10.424\;-\;1.107x\)
\(\displaystyle 4.107x\;=\;10.424\)
\(\displaystyle x\;=\;\frac{10.424}{4.107}\;=\;2.54\;atm\)
\(\displaystyle [CO]_{eq}\;=\;3x\;=\;3\times 2.54\;atm\;=\;\mathbf{7.62\;atm}\;\;\;\;[CO_2]_{eq}\;=\;28.25\;atm\;-\;3\;\times 2.54\;atm\;=\;\mathbf{20.63\;atm}\)
Back to The Simplifying Assumption
Exercise 3. What are the equilibrium concentrations of all species in a 0.45 M NH3 solution?
NH3 (aq) + H2O (l) ⇄ NH4+ (aq) + OH– (aq) Kc = 1.8 x 10-5
Set up ICE table.
\(\displaystyle K_c\;=\;\frac {[NH_4]^+[OH^-]}{[NH_3]}\;=\;1.8 \times 10^{-5}\)
Assume 0.45 M – x ≅ 0.45 M. We will neglect x in the denominator.
\(\displaystyle \frac {x^2}{0.45}\;=\;1.8\times 10^{-5}\)
Solve for x.
\(\displaystyle x\;=\;\sqrt{0.45 \times 1.8\times 10^{-5}}\;=\;0.0030\;M\)
Check assumption. \(\displaystyle \frac {0.0030\;M}{0.45\;M}\times 100\;=\;0.67\%\)
0.67% < 5% Our assumption is valid.
[NH3]eq = 0.45 – x = 0.45 M – 0.0030 M = 0.45 M
[NH3+]eq = [OH–]eq = 0.0030 M
Back to The Simplifying Assumption
Exercise 4. Calculate the equilibrium concentrations of all species if the initial concentration of HCN is 0.25 M.
HCN (aq) ⇄ H+ (aq) + CN– (aq) Kc = 4.9 x 10-10
Set up ICE table.
\(\displaystyle K_c\;=\;\frac {[H^+][CN^-]}{[HCN]}\;=\;4.9 \times 10^{-10}\)
Assume 0.25 M – x ≅ 0.25 M. We will neglect x in the denominator.
\(\displaystyle x\;=\;\sqrt{0.25\times (4.9\times 10^{-10}}\;=\;1.1 \times 10^{-5}\;M\)
Check assumption using 5% rule
\(\displaystyle \frac{1.1 \times 10^{-5}\;M}{0.25\;M}\times 100\;=\;0.004\%\)
Our assumption is valid 0.004% < 5%.
[HCN]eq = 0.25 M – x = 0.25 M – (1.1 x 10-5 M = 0.25 M
[H+]eq = [CN–]eq = 1.1 x 10-5 M
Back to The Simplifying Assumption
Exercise 5. The following reaction has Kc = 4.60 x 10-3 at a certain temperature.
N2O4 (g) ⇄ 2 NO2 (g)
The initial concentration of N2O4 was 0.500 M. What are the concentrations of N2O4 and NO2 at equilibrium?
Set up ICE table.
\(\displaystyle K_c\;=\;\frac{[NO_2]^2}{[N_2O_4]}\;=\;4.60 \times 10^{-3}\)
\(\displaystyle \frac{(2x)^2}{(0.500\;-\;x)}\;=\;4.60 \times 10^{-3}\)
In this problem, the assumption will not work — there will be greater than 5% error. Dividing 0.500 M/(4.6 x 10-3) = 109 which is less than 400. We have to use the quadratic formula.
\(\displaystyle \frac{4x^2}{(0.500\;-\;x)}\;=\;4.60 \times 10^{-3}\)
Solve the equation for x using the quadratic formula. First multiply and then collect all like terms and set the equation to zero.
\(\displaystyle 4x^2\;=\;(0.500-x)\times (4.60\;\times 10^{-3})\)
\(\displaystyle 4x^2\;=\;0.0023 \times (-4.60\;\times 10^{-3}x)\)
\(\displaystyle 4x^2\;+\;(4.60\;\times 10^{-3}x)\;-\;0.0023\;=\;0\)
We have an equation in the form of ax2 + bx + c = 0 where a = 4, b = 4.60 x 10-3, and c = -0.00230.
The quadratic formula is: \(\displaystyle x\;=\;\frac{-b\pm\sqrt{b^2\;-\;4ac}}{2a}\)
\(\displaystyle \frac{-4.60\times 10^{-3}\pm\sqrt{(4.60 \times 10^{-3})^2\;-\;4 \times 4 \times(-0.0230)}}{2 \times 4}\)
\(\displaystyle \frac {-4.60 \times 10^{-3}\pm 0.192}{8}\)
x = 0.0234 M and x = -0.0246 M
x cannot be negative so x = 0.0234 M
The equilibrium concentrations are:
[N_2O_4]eq = 0.500 M – x = 0.500 M – 0.0234 M = 0.477 M
[NO_2]eq = 2x = 2 x 0.0234 M = 0.0468 M