Given equilibrium reaction is endothermic, hence we can write equation as shown below.
3 O 2 (g) + Heat
2 O 3 (g)
Now, by applying Le-chateliars principle we can solve given problem.
According to this principle, when reactant is added to equilibrium reaction or one of the product is removed , equilibrium disturbs and to attain equilibrium reaction proceeds in forward direction.
In this case , oxygen and heat are reactants, hence addition of them shift reaction to the product side ( i e right side)
According to this principle, when product is added to equilibrium reaction or reactant is removed, equilibrium disturbs and to attain equilibrium reaction proceeds in backward direction.
in this case , addition of O 3 or removal of heat will shift reaction to left side.
Catalyst does not affect position of equilibrium
Stress | Equilibrium shift ? |
Increase O 2 concentration | ![]() |
Increase O 3 concentration | ![]() |
Add heat | ![]() |
Catalyst | No Shift |
Cool reaction | ![]() |
PART 2
Consider given reaction , C6H6 (l) + Br
2 (l)
C6H5Br (l) + HBr (g)
From reaction, 1 mol C6H6
1 mol Br 2
1 mol C6H5Br
1 mol HBr
To find limiting reactant we must know molar mass of both reactants.
Molar mass of benzene ( C6H6 ) = ( 6
12.01 ) + ( 6
1.0079 ) = 78.11 g / mol
Molar mass of Br 2 = ( 2
79.91 ) = 159.82 g / mol
We have , 1 mol C6H6
1 mol Br 2
i e 78.11 g C6H6
159.82 g Br 2
40 g C6H6
159.82
40 / 78.11 g Br 2
40 g C6H6
81.84 g Br 2
40 g benzene will require 81.84 g Br 2 . The provided amount of Br 2 is 125 g. Hence, it is excess reactant.
Therefore, benzene is limiting reactant .
ANSWER : Limiting reactant : Benzene
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