Q. What is $\mathrm{\Delta}n$ for combustion of 1 mole of benzene, when both the reactants and products are gas at 298 K
A) 0
B) 1
C) 0.5
D) 1.5
Explanation
(c)
$C_{6}H_{6}(g) + \frac{15}{2}O_{2}(g) \rightarrow 6CO_{2}(g) + 3H_{2}O(g)$
$\mathrm{\Delta}n = 6 + 3 - 1 - \frac{15}{2} = + 0.5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.