Discussion Forum

The combustion of benzene[l]gives CO₂[g] anf 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= 8.314J K^-1mol^-1)

Answer:

Explanation:

Key Idea   Calculate the heat of combustion with the help of following formula

$\triangle H_{P}=\triangle U+\triangle n_{g}RT$

where $\triangle H_{P}$= Heat of combustion at constant pressure

$\triangle U$= Heat at constant volume (It is also called $\triangle E$)

$\triangle n_{g}$ = Change in number of moles (In gaseous state)

R=Gas constant; T= Temperature.

From the equation,

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

Change in the number of gaseous moles i.e.

$\triangle n_{g}=6-\frac{15}{2}=-\frac{3}{2} or-1.5$

Now we have $\triangle n_{g}$ and other values given in the question are

$\triangle U=-3263.9kJ/mol$

  T=25°C =273+25=298K

R=8.314JK^-1mol^-1

So, $\triangle H_{p}=(-3263.9)+(-1.5)\times 8.314\times 10^{-3} \times 298$

= -3267.6 kJmol^-1