Q

# Can someone help me with this, 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 258 C; heat of combustion (in kJ mol−1) of benzene at constant

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 258 C; heat of combustion (in kJ mol−1) of benzene at constant pressure will be :

(R=8.314 JK−1 mol−1)

• Option 1)

−3267.6

• Option 2)

4152.6

• Option 3)

−452.46

• Option 4)

3260

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As we learned

Molar heat capacity for isobaric process C(p) -

$dH=n\: C_{P}\: dT$

- wherein

$dH=dE+d(PV)$

or

$dH=dE+n\, R\, dT$

$C_{6}H_{6}(l)+\frac{15}{2}O_{2}(g)\rightarrow 6CO_{2}(g)+3H_{2}O(l)$

$\Delta ng= 6-\frac{15}{2}$

For chemical reactions;

$\Delta H= \Delta U+\Delta n_{g}\: RT$        $therefore \Delta U= Q_{V} and\: \Delta H= Q_{P}$

$\Delta H= -3263,9\times 1000+\left ( -\frac{3}{2} \right )\times 8.314\times 298$

$\Delta H= \left ( -3263,9\times 1000-3716,36 \right )J= -3267.6kJ$

Option 1)

−3267.6

Option 2)

4152.6

Option 3)

−452.46

Option 4)

3260

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