1. Tardigrade
  2. Question
  3. Chemistry
  4. The combustion of benzene (l) gives CO2(g) and H2O(l) . Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol-1 at 25° C ; heat of combustion (in kJ mol-1 ) of benzene at constant pressure will be - (R = 8.314 JK-1 mol-1)

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Q. The combustion of benzene $ (l)$ gives $CO_2(g)$ and $H_2O(l)$ . Given that heat of combustion of benzene at constant volume is $-3263.9 \, kJ \, mol^{-1}$ at $25^{\circ} C$ ; heat of combustion (in $ kJ \, mol^{-1}$ ) of benzene at constant pressure will be -$ (R = 8.314 \, JK^{-1} \, mol^{-1}) $

JEE MainJEE Main 2018Thermodynamics

Solution:

$C _{6} H _{6}( I )+\frac{15}{2} O _{2}( g ) \longrightarrow 6 CO _{2}( g )+3 H _{2} O ( I )$

$\Delta n _{ g }=6-\frac{15}{2}=-\frac{3}{2}$

$\Delta H =\Delta U +\Delta n _{ g } RT$

$=-3263.9+\left(-\frac{3}{2}\right) \times 8.314 \times 298 \times 10^{-3}$

$=-3263.9+(-3.71)$

$=-3267.6 \,kJ\, mol ^{-1}$