The combustion of benzene l gives CO2g and H2Ol- Given that the heat of combustion of benzene at constant volume is -3263-9 kJ mol-1 at 25- C- The heat of combustion in kJ mol-1 of benzene at constant pressure will be- R-8-314 JK-1mol-1

The heat liberated on complete combustion of $7.8 g$ benzene is $327 K J$ . This heat has been measured at constant volume and at $27^{0} C$ . What is the heat of combustion of benzene at constant pressure at $27^{0} C (R = 8.3 J m o l^{- 1} K^{- 1})$

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The combustion of benzene (l) gives CO2(g) and H2O(l). Given that the heat of combustion of benzene at constant volume is —3263.9 kJ mol−1 at 25∘C. The heat of combustion (in kJ mol−1) of benzene at constant pressure will be: (R=8.314 JK−1mol−1)

The correct option is A -3267.6C6H6(l)+152O2(g)→6CO2(g)+3H2O(l) Here, Δng=6−152=−32 We know, ΔH=ΔU+ΔngRT ΔH=(−3263900−32×8.314×298) J =−3267616 J=−3267.6 kJ

The correct option is A -3267.6
$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$
Here, $Δ n_{g} = 6 - \frac{15}{2} = - \frac{3}{2}$
We know, $Δ H = Δ U + Δ n_{g} R T$
$Δ H = (- 3263900 - \frac{3}{2} \times 8.314 \times 298) J$
$= - 3267616 J = - 3267.6 k J$