Question

The Heat Of Combustion Of Benzene At $270°\mathrm{C}$ Found By Bomb Calorimeter I.E. For The Reaction ${\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)+7\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to 6{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ Is $780\mathrm{K}.\mathrm{Cal}{\mathrm{mol}}^{-1}$. The Heat Evolved On Burning $39\mathrm{g}$ Of Benzene In An Open Vessel Will Be

Answer 1

Heat of combustion:

• The term "heat of combustion" refers to the amount of heat needed to totally burn down a chemical species when oxygen is present.

Step 1: Finding change in internal energy

• ${\mathrm{C}}_6{\mathrm{H}}_6$ has a molar mass of = $6\times 12+6\times 1=78\mathrm{grams}$.
• The energy content of $78\mathrm{grams}$of benzene is $780\mathrm{K}.\mathrm{Cal}{\mathrm{mol}}^{-1}$
• Energy for $39\mathrm{g}$of benzene is equal to $\frac{780\times 39}{78}=390.$
• Therefore, $∆\mathrm{U}=390\mathrm{K}.\mathrm{Cal}{\mathrm{mol}}^{-1}$

Step 2: Finding change in number of moles of gaseous species

• $$\begin{gathered} \mathrm{T}=27+273\mathrm{K} \\ \mathrm{T}=300\mathrm{K} \end{gathered}$$
• The indicated response is
• ${\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)+7\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to 6{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
• Change in mole number (n)=$6-7.5=-1.5$

Step 3: Finding change in heat

• $\mathrm{R}=2\times {10}^{-3}\mathrm{Kcal}$
• Having said that,
• $$\begin{gathered} \mathrm{\Delta H}=\mathrm{\Delta U}+\mathrm{\Delta nRT} \\ \mathrm{\Delta H}=390+(-1.5\times 2\times {10}^{-3}\times 300) \\ \mathrm{\Delta H}=390-0.9 \\ \mathrm{\Delta H}=389.1\mathrm{Kcal} \end{gathered}$$

$39\mathrm{g}$ of benzene burned in an open vessel will produce $389.1\mathrm{Kcal}$ of heat.