3.3 The equilibrium constant

<< Chapter < Page

Siyavula textbooks: grade 12 Page 2 / 2

Chapter >> Page >

For the reaction:

$S O_{2} (g) + N O_{2} (g) \to N O (g) + S O_{3} (g)$

the concentration of the reagents is as follows:

[SO $_{3}$ ] = 0.2 mol.dm $^{- 3}$

[NO $_{2}$ ] = 0.1 mol.dm $^{- 3}$

[NO] = 0.4 mol.dm $^{- 3}$

[SO $_{2}$ ] = 0.2 mol.dm $^{- 3}$

Calculate the value of K $_{c}$ .

Write the equation for k $_{c}$
$K_{c} = \frac{[N O] [S O_{3}]}{[S O_{2}] [N O_{2}]}$
Fill in the values you know for this equation and calculate K $_{c}$
$K_{c} = \frac{(0.4 \times 0.2)}{(0.2 \times 0.1)} = 4$

Got questions? Get instant answers now!

For the reaction:

$S (s) + O_{2} (g) ⇌ S O_{2} (g)$

Write an equation for the equilibrium constant.
Calculate the equilibrium concentration of $O_{2}$ if K $_{c}$ =6 and [ $S O_{2}] = 3 m o l . d m^{- 3}$ at equilibrium.

Write the equation for K $_{c}$
$K_{c} = \frac{[S O_{2}]}{[O_{2}]}$

(Sulfur is left out of the equation because it is a solid and its concentration stays constant during the reaction)
Re-arrange the equation so that oxygen is on its own on one side of the equation
$[O_{2}] = \frac{[S O_{2}]}{K_{c}}$
Fill in the values you know and calculate [O $_{2}$ ]
$[O_{2}] = \frac{3 m o l . d m^{- 3}}{6} = 0.5 m o l . d m^{- 3}$

Got questions? Get instant answers now!

Initially 1.4 moles of NH $_{3}$ (g) is introduced into a sealed 2.0 dm $^{- 3}$ reaction vessel. The ammonia decomposes when the temperature is increased to 600K and reaches equilibrium as follows:

$2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$

When the equilibrium mixture is analysed, the concentration of NH $_{3}$ (g) is 0.3 mol $\cdot$ dm $^{- 3}$

Calculate the concentration of N $_{2}$ (g) and H $_{2}$ (g) in the equilibrium mixture.
Calculate the equilibrium constant for the reaction at 900 K.

Calculate the number of moles of NH $_{3}$ at equilibrium.
$c = \frac{n}{V}$

Therefore,

$n = c \times V = 0.3 \times 2 = 0.6 m o l$
Calculate the number of moles of ammonia that react (are 'used up') in the reaction.
Moles used up = 1.4 - 0.6 = 0.8 moles
Calculate the number of moles of product that are formed.
Remember to use the mole ratio of reactants to products to do this. In this case, the ratio of NH $_{3}$ :N $_{2}$ :H $_{2}$ = 2:1:3. Therefore, if 0.8 moles of ammonia are used up in the reaction, then 0.4 moles of nitrogen are produced and 1.2 moles of hydrogen are produced.
Complete the following table

NH $_{3}$ N $_{2}$ H $_{2}$

Start of reaction 1.4 0 0

Used up 0.8 0 0

Produced 0 0.4 1.2

Equilibrium 0.6 0.4 1.2
Using the values in the table, calculate [N $_{2}$ ] and [H $_{2}$ ]
$[N_{2}] = \frac{n}{V} = \frac{0.4}{2} = 0.2 m o l . d m^{- 3}$

$[H_{2}] = \frac{n}{V} = \frac{1.2}{2} = 0.6 m o l . d m^{- 3}$
Calculate K $_{c}$
$K_{c} = \frac{{[H_{2}]}^{3} [N_{2}]}{{[N H_{3}]}^{2}} = \frac{{(0.6)}^{3} (0.2)}{{(0.3)}^{2}} = 0.48$

Got questions? Get instant answers now!

Hydrogen and iodine gas react according to the following equation:

$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$

When 0.496 mol $H_{2}$ and 0.181 mol $I_{2}$ are heated at $450^{o}$ C in a 1 dm $^{3}$ container, the equilibrium mixture is found to contain 0.00749 mol $I_{2}$ . Calculate the equilibrium constant for the reaction at $450^{o}$ C.

Calculate the number of moles of iodine used in the reaction.
Moles of iodine used = 0.181 - 0.00749 = 0.1735 mol
Calculate the number of moles of hydrogen that are used up in the reaction.
The mole ratio of hydrogen:iodine = 1:1, therefore 0.1735 moles of hydrogen must also be used up in the reaction.

Calculate the number of moles of hydrogen iodide that are produced.

The mole ratio of H $_{2}$ :I $_{2}$ :HI = 1:1:2, therefore the number of moles of HI produced is 0.1735 $\times$ 2 = 0.347 mol.

So far, the table can be filled in as follows:

	H $_{2}$ (g)	I $_{2}$	2HI
Start of reaction	0.496	0.181	0
Used up	0.1735	0.1735	0
Produced	0	0	0.347
Equilibrium	0.3225	0.0075	0.347

Calculate the concentration of each of the reactants and products at equilibrium.
$c = \frac{n}{V}$

Therefore the equilibrium concentrations are as follows:

[H $_{2}$ ] = 0.3225 mol.dm $^{- 3}$

[I $_{2}$ ] = 0.0075 mol.dm $^{- 3}$

[HI] = 0.347 mol.dm $^{- 3}$
Calculate K $_{c}$
$K_{c} = \frac{[H I]}{[H_{2}] [I_{2}]} = \frac{0.347}{0.3225 \times 0.0075} = 143.47$

Got questions? Get instant answers now!

The equilibrium constant

Write the equilibrium constant expression, K c for the following reactions:
1. $2 NO (g) + {Cl}_{2} (g) ⇌ 2 NOCl$
2. $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$
The following reaction takes place: ${Fe}^{3 +} (aq) + 4 {Cl}^{-} ⇌ {FeCl}_{4}^{-} (aq)$ K $_{c}$ for the reaction is 7.5 $\times$ 10 $^{- 2}$ mol.dm $^{- 3}$ . At equilibrium, the concentration of FeCl $_{4}^{-}$ is 0.95 $\times$ 10 $^{- 4}$ mol.dm $^{- 3}$ and the concentration of free iron (Fe $^{3 +}$ ) is 0.2 mol.dm $^{- 3}$ . Calculate the concentration of chloride ions at equilibrium.
Ethanoic acid (CH $_{3}$ COOH) reacts with ethanol (CH $_{3}$ CH $_{2}$ OH) to produce ethyl ethanoate and water. The reaction is: ${CH}_{3} COOH + {CH}_{3} {CH}_{2} OH \to {CH}_{3} {COOCH}_{2} {CH}_{3} + H_{2} O$ At the beginning of the reaction, there are 0.5 mols of ethanoic acid and 0.5 mols of ethanol. At equilibrium, 0.3 mols of ethanoic acid was left unreacted. The volume of the reaction container is 2 dm $^{3}$ . Calculate the value of K $_{c}$ .

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Siyavula textbooks: grade 12 physical science. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11244/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Siyavula textbooks: grade 12 physical science' conversation and receive update notifications?

Ask

	15 Dr. Amberg Pharm quiz By Brooke Delaney Start Exam
	Lean Startup Quiz By Yasser Ibrahim Start Quiz
©flickr:	Morphology By Jugnu Khan Start Quiz
	Grade 10 Module 2.1 IT Quiz (Part 1) By Christine Zeelie Start Quiz
	Algebra and trigonometry By OpenStax Read Online Course
	NCE Ch 03 Human Growth and Develoment By Anh Dao Start Test
	Thermal-Fluid Systems MCQ By Steve Gibbs Start Quiz
	2 Human A&P 2- Test 2 By Madison Christian Start Flashcards
	Anthropology Religion Culture By Richley Crapo Start Assignment
	20 Biology 20 Phylogenies the History of Life MCQ By OpenStax Start Quiz

	NH $_{3}$	N $_{2}$	H $_{2}$
Start of reaction	1.4	0	0
Used up	0.8	0	0
Produced	0	0.4	1.2
Equilibrium	0.6	0.4	1.2