Question

0.16 g of methane was subjected to combustion at ${27}^{\circ }$ C in a bomb Calorimeter. The temperature of calorimeter system (including water) was found to rise by ${0.5}^{\circ }$ C. The heat of combustion of methane at :

( i ) constant volume and ( ii ) constant pressure.

[Given: The thermal capacity of calorimeter system is 17.7 kJ $K^{-1}.$ $(R=8.314Jmo{l}^{-1}{K}^{-1})]$

• A. ( i ) $-$885 kJ / mol ( ii ) $-$889.95 kJ / mol
• B. ( i ) $-$785 kJ / mol ( ii ) $-$859.95 kJ / mol
• C. ( i ) $-$587kJ / mol ( ii ) $-$789.95 kJ / mol
• D. ( i ) $-$985 kJ / mol ( ii ) $-$999.95 kJ / mol

Answer 1

Answer: (A)

( i ) $-$885 kJ / mol ( ii ) $-$889.95 kJ / mol

${CH}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)$

Heat of combustion at constant volume i.e., $q_v=\Delta E=$ internal energy.

$\Delta E=$ Heat capacity of calorimeter$\times$(rise in temperature)$\times$((molar mass of compound)/(mass of compound))

$=17.7\times 0.5\times \frac{16}{0.16}=885$

$q_v=\Delta E=-885kJ/mol$

Heat of combustion at constant pressure $=q_p=\Delta H=$ethalpy

$\Delta H=\Delta E+\Delta nRT$

$\Delta n=1-3=-2$

$T=300K;R=8.314\times {10}^{-3}kJ.K^{-1}/mol$

So, $\Delta H=-885+(-2)\times (8.314\times {10}^{-3})\times 300=-889.99kJ/mol$