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Calculating an Equilibrium Concentration from the Equilibrium Constant

1. Introduction
2. The ICE Table
3. Problems
4. Solutions
5. References
6. Outside Links
7. Contributors

To calculate an equilibrium concentration from an equilibrium constant, first we need to have prior knowledge of what equilibrium is and how to write an equilibrium constant. Hence, equilibrium is a state of dynamic balance where the change in reactant and product populations are ratio.

1. Introduction
2. The ICE Table
3. Problems
4. Solutions
5. References
6. Outside Links
7. Contributors

Introduction

An equilibrium constant is the ratio of the equilibrium constant expression. It is the ratio of the concentration of the products to reactants.

$\ aA + bB {\rightleftharpoons} \ cC + dD$

$K_{c} = \dfrac{[A]^{a}[B]^{b}}{[C]^{c}[D]^{d}}$

The ICE Table

The easiest approach to this problem is to use the ICE Table. The ICE Table is an organized way to determine what givens we have and what we need to find. The "I" stands for initial which is the initial amount given, "C" represents change which is the change that happened, and "E" is for equilibrium which is the final step.

Example

Given: At equilibrium for the gaseous reaction what is the concentration for each substance?

$\ C_{2}H_{4} + H_{2} {\rightleftharpoons} C_{2}H_{6} $ $K_{c} = 0.98 $

$C_{2}H_{4} = 0.33 $ $H_{2} = 0.53 $

Solution:

$K_{c} = \dfrac{[C_{2}H_{6}]}{[C_{2}H_{4}][H_{2}]}$

$0.98= \dfrac{[C_{2}H_{6}]}{[C_{2}H_{4}][H_{2}]}$

ICE Table

C₂H₆ is not mentioned, so its value is 0, meaning its change must be increasing.

	$C_{2}H_{4} $	$H_{2} $	$C_{2}H_{6} $
Initial	0.33	0.53	0
Change
Equilibrium

Because ethane, C₂H₆is a product, and we cannot have negative concentrations, the reactants are decreasing. To represent this, we use the x value and a positive or negative sign in front in the ICE table.

	$C_{2}H_{4} $	$H_{2} $	$C_{2}H_{6} $
Initial	0.33	0.53	0
Change	-x	-x	+x
Equilibrium

Equilibrium is determined by adding Initial and Change together.

	$C_{2}H_{4} $	$H_{2} $	$C_{2}H_{6} $
Initial	0.33	0.53	0
Change	-x	-x	+x
Equilibrium	0.33-x	0.53-x	x

Now if we substitute what we have in the ICE table into the equilibrium constant expression, we will have the value of the equilibrium concentration.

$0.98= \dfrac{x}{(0.33-x)(0.53-x)}$

$0.98= \dfrac{x}{(x^{2} + 0.86x + 0.1749)}$

$0.98 {(x^{2} + 0.86x + 0.1749)}= {x}$

$0.98x^{2} + 0.8428x + 0.171402= x$

$0.98x^{2} - 1.8428x + 0.171402= 0$

Now using the quadratic formula we can solve for x.

$ x= \dfrac{-b + {[{b^2} - 4ac]^{\dfrac{1}{2}}}}{2a}$

$ x= \dfrac{1.8428 + {[{1.8428^2} - 4(0.98)(0.171402)]^{\dfrac{1}{2}}}}{2(0.98)}$

$x = 1.7823 or 0.0981$

Now we plug in x to determine which x is false.

$ C_{2}H_{4} = (0.33-1.7823) = -1.4523$

$ C_{2}H_{4} = (0.33-0.0981) = 0.2319$

If x = -1.7823 then x is negative meaning the concentration is negative, therefore, x must equal 0.0981.

So:

$ C_{2}H_{4} = 0.2319M$

$ H_{2} = (.53-.0981) = 0.4319$

$ C_{2}H_{6} = 0.0981$

Problems

1. Find the concentration of iodine in the following reaction when the equilibrium constant is 3.76 X 10³. 2 mol of iodine is placed in a 2 L flask at 100K.

$\ I_{2}(g) {\rightleftharpoons} 2I(aq)$

2. What is the concentration of silver ion in 1.00 L of solution with 0.020 mol of AgCl and 0.020 mole of Cl^- with the following reaction? The equilibrium constant is 1.8 X 10^-10.

$\ AgCl(s) {\rightleftharpoons} Ag^{+}(aq) + Cl^{-}(aq)$

3. What are the concentrations of the products and reaction for the following equilibrium reaction?

Given Initials:$\ HSO_{4}^{-} = 0.4 $ $H_{3}O^{+} = 0.01$ $SO_{4}^{2-}= 0.07 $ $K=.012 $

$\ HSO_{4}^{-}(aq) + H_{2}O(l) {\rightleftharpoons} H_{3}O^{+}(aq) + SO_{4}^{2-}(aq) $

4. The initial concentration of HCO₃ is 0.16 M in the following reaction. What is the H⁺ concentration at equilibrium? K_c=0.20.

$\ H_{2}CO_{3} {\rightleftharpoons} H^{+}(aq) + CO_{3}^{2-}(aq)$

5.The initial concentration of PCl₅ is 0.200 moles per liter and there is no products in the system when the reaction starts. If the equilibrium constant is 0.030, calculate all the concentrations at equilibrium.

Solutions

	$I_{2}$	$I$
Initial	2mol/2L = 1 M	0
Change	-x	+2x
Equilibrium	1-x	2x

At equlibrium

$K_c=\dfrac{[I]^2}{[I_2]}$

$3.76x10^3=\dfrac{(2x)^2}{1-x} = \dfrac{4x^2}{1-x}$

cross multiply

$4x^2+3.76.10^3x-3.76.10^3=0$

apply the quadratic formula:

$ \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

with: a=4, b=3.76x10³ c=-3.76x10³

with solutions of x=0.999 and -940. The latter solution is unphysical (a negative concentration). So x=0.999 at equilibrium.

[I]=2x=1.99 M

[I2]=1-x=1-.999=.001M

	$Ag^{+} $	$Cl^{-} $
Initial	0	0.02mol/1.00 L = 0.02 M
Change	+x	+x
Equilibrium	x	0.02 + x

$K_{c}=[Ag^{-}][Cl^-]$

$1.8x10^{-10}=(x)(.02+x)$

$x^2+.02x-1.8x10^{-10}=0$

$x = 9.00 X 10^{-9}$

[Ag^-]=x=9.00x10^-9

[Cl^-]=0.02+x=0.020

	$H_{2}CO_{3} $	$SO_{4}^{2-} $	$H_{3}O $
Initial	0.4	0.01	0.07
Change	-x	+x	+x
Equilibrium	0.4-x	0.01+x	0.07+x

$K_{c}=[SO_{4}^{2-}][H_{3}O]/[H_{2}CO_{3}]$

$.012=(.01+x)(.07+x)/(.4-x)$

cross multiply and get:

$x^{2}+.2x-.0041=0$

apply the quadratic formula

x = 0.0328

[H₂CO₃]=0.4-x=0.4-0.0328=0.3672

[S0₄^2-]=0.01+x=0.01+0.0328=0.0428

[H₃0]=0.07+x=0.07+0.0328=0.1028

	H₂CO₃	$H^{+} $	$CO_{3}^{2-} $
Initial	.16	0	0
Change	-x	+x	+x
Equilibrium	.16-x	x	x

$K_{c}=[CO_{3}^{2-}][H^{+}]/[H_{2}CO_{3}]$

$.20=(x)(x)/.16-x$

$x^{2}+.20x-.032=0$

apply the quadratic equation

x=0.1049

[H⁺]=x=0.1049

5. First write out the balanced equation:

$\ PCl_{5}(g) {\rightleftharpoons} PCl_{3}(g) + Cl_{2}(g)$

	$PCl_{5} $	$PCl_{3} $	$Cl_{2} $
Initial	0.2	0	0
Change	-x	+x	+x
Equilibrium	0.2-x	x	x

$K_{c}=[PCl_{3}][Cl_{2}]/[PCl_{5}]$

$.030=(x)(x)/(.2-x)$

Cross multiply

$x^{2}+.030x-.006=0$

apply quadratic

x=0.064

[PCl₅]=0.2-x=0.136

[PCl₃]=0.064

[Cl₂]=0.064

References

Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2007.
Criddle, Craig and Larry Gonick. The Cartoon Guide to Chemistry. New York: HarperCollins Publishers, 2005.

Outside Links

http://chemed.chem.purdue.edu/genche...6/equilib.html

Contributors

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