MCQ
What is $\Delta n$ for combustion of $ 1 $ mole of benzene, when both the reactants and the products are gas at $298\, K$
- A$0$
- B$ + 3/2$
- C$-3/2$
- ✓$+ 1/2$
✓
Answer
Correct option: D.
$+ 1/2$
(d)${C_6}{H_{6(g)}} + \frac{{15}}{2}{O_{2(g)}} \to 6C{O_{2(g)}} + 3{H_2}{O_{(g)}}$
$\Delta n = 6 + 3 - 1 - \frac{{15}}{2} = + \frac{1}{2}$.
$\Delta n = 6 + 3 - 1 - \frac{{15}}{2} = + \frac{1}{2}$.
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