Question

A sample of $C{H}_4$ of 0.08 g was subjected to combustion at ${27}^{o}C$ in a bomb calorimeter. The temperature of the calorimeter system was found to be raised by ${0.25}^{o}C$. If heat capacity of calorimeter is 18 kJ, $\Delta H$ for combustion of $C{H}_4$ at ${27}^{o}C$ is:

• A. $-900$ kJ/mole
• B. $-905$ kJ/mole
• C. $-895$ kJ/mole
• D. $-890$ kJ/mole

Answer 1

The correct option is B $-905$ kJ/mole

${CH}_4\left(g\right)+{2O}_2\left(g\right)⟶{CO}_2\left(g\right)+2H_{2}O\left(l\right)$

$\Delta E=$ Heat of combustion

$=$ Heat capacity $\times$ rise in $T$ $\times$ $\frac{Molarmass}{Massofcompound}$

$=-18\times 0.25\times \frac{16}{0.08}$

$=-900$ KJ/mole $R=8.314\times {10}^{-3}KJ{mol}^{-1}$

$\Delta H=\Delta E+\Delta nRT$ $\Delta n=1-3=-2$

$=-900+(-2)\times 8.314\times {10}^{-3}\times 300$ $T=300K$

$=-900-4.9884=-904.98\simeq -905KJ/mol$