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

C_6H_6(l)+15/2O_2(g)→ 6CO_2(g)+3H_2O(l) Here, Δ n_g=6 15/2= 3/2 We know, Δ H=Δ U+Δ n_gRT Δ H= ( 3263900 3/2× 8.314×298) J nbsp; nbsp; nbsp; nbsp; ...

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Standard XII

Chemistry

Limitations of First Law

The combustio...

Question

The combustion of benzene $(l)$ gives $C O_{2} (g)$ and $H_{2} O (l)$ . Given that the heat of combustion of benzene at constant volume is —3263.9 $k J m o l^{- 1}$ at $25^{\circ} C$ . The heat of combustion $(i n k J m o l^{- 1})$ of benzene at constant pressure will be:
$(R = 8.314 J K^{- 1} m o l^{- 1})$

-3267.6

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Solution

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, $Δ H = Δ U + Δ n_{g} R T$
$Δ H = (- 3263900 - \frac{3}{2} \times 8.314 \times 298) J$
$= - 3267616 J = - 3267.6 k J$

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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, $Δ H = Δ U + Δ n_{g} R T$
$Δ H = (- 3263900 - \frac{3}{2} \times 8.314 \times 298) J$
$= - 3267616 J = - 3267.6 k J$