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JulijaS [17]
3 years ago
15

Bromine (Br,) is observed as a red-brown color During the bromination of benzene, an electrophilic aromatic substitution, why do

es the red-brown color of Br2 disappear as the reaction progresses?
a. The bromine (Br2) dissolves in the solution and becomes colorless
b. The bromine (Br2) has reacted with the benzene, leaving fewer bromine molecules in solution.
c. The bromine (Br,) reacts with acid to produce HBr, which is colorless .
d. The bromine (Br,) leaves the mixture as bromine gas .
Chemistry
1 answer:
sergey [27]3 years ago
4 0

Answer:

b. The bromine (Br2) has reacted with the benzene, leaving fewer bromine molecules in solution.

Explanation:

Colour is caused by absorption of light of the frequency of “complementary colours”. Exactly which frequencies molecules absorb depends on the arrangement of electrons. Obviously, to break the Br-Br bond and the C=C bond in the reactants and replace them with two C-H bonds is a significant reordering of electrons and so the new molecule is no longer able to absorb the same frequencies as previously. Hence, a colour change, in this case from orange to colourless.

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Explanation:

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1. N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

As 4 moles of gaseous reactants are changing to 2 moles of gaseous products,  the randomness is decreasing and the entropy is negative

2. 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

As 9 moles of gaseous reactants are changing to 10 moles of gaseous products,  the randomness is increasing and the entropy is positive.

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As 1 mole of solid reactants is changing to 2 moles of gaseous products, the randomness is increasing and the entropy is positive.

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As 4 moles of gaseous reactants is changing to 5 moles of gaseous products, the randomness is increasing and the entropy is positive

5. 2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(l)

As 4 moles of gaseous reactants is changing to 1 moles of gaseous products, the randomness is decreasing and the entropy is negative.

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