Question

0.16 g of methane was subjected to combustion at ${27}^{\circ }C$ in bomb calorimeter. The temperature of calorimeter system (including water) was found to rise by ${0.5}^{\circ }C$. If the heat of combustion of methane at constant volume and constant pressure is $x$ kJ/mole, find the value of $x$. The thermal capacity of the calorimeter system is $17.7kJ{K}^{-1},R=8.314J{K}^{-1}mo{l}^{-1}$.

• A. $-890$
• B. $890$
• C. $450$
• D. $-450$

Answer 1

The correct option is A $-890$

Heat given by burning of 0.16 g methane = heat taken up by calorimeter = $17.7\times 0.5=8.85kJ$

$\therefore$ Heat given by burning of 16 g methane

$=\frac{8.85\times 16}{0.16}=885kJ$

The burning is taking place at constant volume in a bomb calorimeter.

$\therefore \Delta U=-885kJmo{l}^{-1}$

$\because \Delta H=\Delta U+\Delta nRT$

$C{H}_4+2O_2⟶C{O}_2+2H_{2}O\left(l\right)$;

$\therefore \Delta n=-2$

$\therefore \Delta H=-885+(-2)\times 8.314\times {10}^{-3}\times 300$

$=-889.98kJmo{l}^{-1}$

Hence, the correct option is $\text{A}$