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Komok [63]
3 years ago
7

Match the effects on the concentration of SO3 gas when the following changes occur after initial equilibrium has been establishe

d in this system (N.C. = no change). 2SO2(g) + O2(g) 2SO3(g) + 46.8 kcal pressure increases increase [O2] increase volume of chamber adding a catalyst raising the temperature
Chemistry
2 answers:
Archy [21]3 years ago
8 0

The reaction is

2SO_{2} (g)+O_{2}(g)->2SO_{3} (g)+46.8 kcal

1) the system is homogeneous with all the reactants and products are gases

2) there are three moles on reactant side and two moles on product side

i) If we increase pressure then there will be decrease in volume. It will cause increase in moles per unit volume. So if we increase pressure the equilibrium will shift towards side having less number of moles of gaseous species.

Thus system will move in forward direction.

ii) If we increase concentration of oxygen it means we are increasing concentration of reactants,this will cause system to shift in forward direction

iii) If we increase volume of chamber,  it will cause decrease in moles per unit volume, hence the system will move in the direction where number of moles of gaseous species is more, System will shift in reverse direction

iv) There is generally no effect of catalyst on equilibrium. However an increase in temperature will cause shift in reverse direction.

mestny [16]3 years ago
3 0

The effects on the concentration of SO3 gas when the following changes occur after initial equilibrium has been established in this system (N.C. = no change) by adding a catalyst.

2SO2(g) + O2(g)  2SO3(g) + 46.8 kcal

So that even though there is an addition of catalyst, no change in reactants or products has occurred because catalyst only provides a faster pathway for the reaction to occur.

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When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
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When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. <u>This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.</u>

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

<u>No of moles = mass of the substance÷molar mass of the substance</u>

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

<u>Thus 52.8 g of CO2 is produced.</u>

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