Thermodynamics Question 9
Question 9 - 01 February - Shift 2
$0.3 g$ of ethane undergoes combustion at $27^{\circ} C$ in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by $0.5^{\circ} C$. The heat evolved during combustion of ethane at constant pressure is $kJ mol^{-1}$.
(Nearest integer)
[Given : The heat capacity of the calorimeter system is $20 kJ K^{-1}, R=8.3 JK^{-1} mol^{-1}$.
Assume ideal gas behaviour.
Atomic mass of $C$ and $H$ are 12 and $1 g mol^{-1}$ respectively]
Show Answer
Answer: (1006)
Solution:
Formula: Application of First Law
(Bomb calorimeter $\rightarrow$ const volume)
Heat released
By combustion of 1 mole
$C_2 H_6(\Delta U)=-\frac{20 \times 0.5}{0.3} \times 30=-1000 kJ$
$C_2 H_6(g)+7 / 2 O_2(g) \to 2 CO_2(g)+3 H_2 O(l)$
$\Delta ng=2-(2+7 / 2)=-(7 / 2)$
$\Delta H=\Delta U+\Delta nRT$
$=-1000-7 / 2 \times 8.3 \times 300 kJ$
$=-1000-6.225$
$=-1006 kJ$
So heat released $=1006 kJ mol^{-1}$
