What is $\Delta \mathrm{n}_{\mathrm{gas}}$ for the combustion of 1 mole of benzene, when both the reactants…

What is $\Delta \mathrm{n}_{\mathrm{gas}}$ for the combustion of 1 mole of benzene, when both the reactants and the products are at $298 \mathrm{~K}$ ?
  1. 0
  2. $1 / 2$
  3. $3 / 2$
  4. $-3 / 2$

Solution

$\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l})+\frac{15}{2} \mathrm{O}_{2}(\mathrm{~g}) ightarrow$
$.6 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}$ (l)
$\Delta \mathrm{n}=6-15 / 2=-3 / 2$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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