Chemical Thermodynamics Question 284

Question: What is $ \Delta n $ for combustion of 1 mole of benzene, when both the reactants and the products are gas at 298 K [Pb. PMT 2001]

Options:

A) 0

B) 3/2

C) - 3/2

D) ½

Show Answer

Answer:

Correct Answer: D

Solution:

$ C_6{H_{6(g)}}+\frac{15}{2}{O_{2(g)}}\to 6C{O_{2(g)}}+3H_2{O_{(g)}} $

$ \Delta n=6+3-1-\frac{15}{2}=+\frac{1}{2} $ .



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