Chemical Thermodynamics Question 120
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} $ .
BETA
