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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

 
Answers (1)
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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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