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Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3
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Rate of chemical Reaction

 The rate of a chemical reaction is the number of moles of reactants converted or products formed per unit time. It varies from one chemical reaction to another because some chemical reactions are faster than others. The rate at which a reaction occurs and its control are significant in industries because they are the factors that determine if the reaction will make economic sense or not.

How to measure the rate of chemical reaction

Consider a reaction between irons and dilute hydrochloric acid

Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)

As the reaction proceeds, the iron is used up and iron (II) chloride and hydrogen gas are formed.

  The rate of the chemical reaction between iron and the acid can be determined by:

·         Taking a given mass of iron

·         Adding an excess of hydrochloric acid

·         Noting the time taken for all the iron to disappear or react

The rate of the chemical reaction can be calculated by:

Rate of chemical reaction = amount of iron (mole)/time taken

Or

Rate of chemical reaction = change in concentration of a reactant or product/time taken

Or

Rate of chemical reaction = mass of reactant/ time taken for the reaction

Example Questions  

1)   When 20g of magnesium trioxocarbonate (IV) oxide was added to excess dilute hydrochloric acid, carbon (IV) oxide evolved. The entire reaction took 10 minutes. Calculate the rate of the reaction.

solution

Rate of reaction = mass of reactant/time taken for the reaction

Mass of the reactant = 20g

Time taken =     10 minutes

Rate of reaction =   20/10 = 2g/min

Or

Mass of reactant = 20g

Time taken = 10x60 = 600s

Rate of reaction = 20/600 = 0.03g/s

2)   If the rate of chemical of calcium trioxocarbonate(IV) with excess hydrochloric acid is 0.2g/s and the mass of calcium trioxocarbonate (IV) which reacted with the acid is 5g, calculate the time it took for the reaction to be completed in minutes.

Solution

Rate of reaction = 0.2g/s

Mass of reactant = 5g

Time taken for the reaction=??

Time taken = 5/0.2 = 25s

In minutes

60s = 1minute

25s = 25/60   = 0.42minutes

Factors Affecting Rate of Reaction

 

 

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Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3
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