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  • Calculate the heat of combustion (in kJ/mol) of benzene at constant pressure, when the heat of combustion of benzene in a bomb calorimeter was found to be   <img alt="\mathrm{-3263.9

Calculate the heat of combustion (in kJ/mol) of benzene at constant pressure, when the heat of combustion of benzene in a bomb calorimeter was found to be   \mathrm{-3263.9\:\: kJ\: mol^{-1}}  at \mathrm{250^{0}C}.

Option: 1

3263.9


Option: 2

-3263.9


Option: 3

3267.6


Option: 4

-3267.6


Answers (1)

best_answer

A bomb calorimeter is a constant volume calorimeter.

The reaction in

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

In this reaction, \mathrm{O_{2}} is the only gaseous reactant and \mathrm{CO_{2}} is the only gaseous product.

\therefore \mathrm{\Delta n_{g}=\sum n_{P}-\sum n_{R}=6-\frac{15}{2}=-\frac{3}{2}}

Also, we are given 

\\\mathrm{\Delta E= -3263.9\:\:kJ\:mol^{-1}} \\\\\mathrm{T=25^{0}C=298K,R=8.314JK^{-1}mol^{-1}} \\\\\mathrm{\Delta H=\Delta E+(\Delta n_{g})RT} \\\\\mathrm{\Delta H=-3263.9+(-3/2)(8.314/1000)(298)} \\\\ \mathrm{\Delta H=-3263.9-3.7=-3267.6\:kJmol^{-1}}

Posted by

jitender.kumar

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