Question

The heat liberated on complete combustion of $7.8g$ benzene is $327KJ$. 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 ${27}^{0}C(R=8.3Jmo{l}^{-1}{K}^{-1})$

• A. -3273.735 KJ
• B. -3274.735 KJ
• C. -3275.735 KJ
• D. -3270.735 KJ

Answer 1

The correct option is A

-3273.735 KJ

Let us first write the chemical reaction which is

$C_6{H}_6\left(l\right)+\frac{15}{2}{O}_2\left(g\right)\to 6C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

$\Delta$ng = number of moles of gaseous products - number of moles of gaseous reactants

$\Delta$ng = $6-\frac{15}{2}$ = $-\frac{3}{2}$

Now $\Delta U$ per mole = $\frac{-327\times 78}{7.8}=-3270KJ$

$\Delta H=\Delta U+\Delta nRT=-3270+(-\frac{3}{2})\left(8.3\right)\left(300\right){10}^{-3}$

= $3270+(-\frac{3}{2})\left(8.3\right)\left(300\right){10}^{-3}$

$\Delta H=-3273.735KJ$