A $1.250 g$ sample of octane $(C_{8} H_{18})$ is burned in excess of oxygen in a bomb calorimeter. The temperature of calorimeter rises from $294.05 K$ to $300.78 K$ . If heat capacity of calorimeter and its contents is $8.93 k J / K$ , calculate the heat transferred to calorimeter.

Open in App

Solution

$C_{8} H_{18} + \frac{25}{2} O_{2} \to 8 C O_{2} + 9 H_{2} O$

Mass of octane, $m = 1.25 g$ or $0.00125 k g$

Heat capacity, $C = 8.93 k J K^{- 1}$

Change in temperature, $Δ T = 300.78 - 294.05 = 6.73 K$

Heat transferred, $Q = m C Δ T$

$= 0.00125 \times 8.93 \times 6.73$

$= 0.075 k J$

Enthalpy of combustion is the enthalpy of reaction. In reaction $1$ mole of octane is used So, enthalpy of $1$ mole of octane can be calculated as follows:

For $1.25 g$ heat transferred is $0.075 k J$

For $1$ mole or $114 g$ of octane heat transferred will be:

$= \frac{0.075}{1.25} \times 114$

$= 6.84 k J$

Suggest Corrections

Similar questions

1.0 g

C_{6} H_{12} O_{6} (s)

{1.56}^{o} C

10.0 k J /^{o} C

298 K

600 J K^{- 1}

0.16 g C H_{4}

27^{o} C

{0.5}^{o} C

17.0 k J K^{- 1}

R = 8.314 J K^{- 1} m o l^{- 1}

C (g r a p h i t e) + O_{2} (g) \to {C O}_{2} (g)

For gas A in a calorimeter heat evolved is 250 kJ/mol. For 0.2 mol of A, temperature rise from 298 K and 300 K. Find out heat capacity of calorimeter

Join BYJU'S Learning Program