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Standard VIII

Chemistry

Coal

The enthalpy ...

Question

The enthalpy of combustion of graphite is – $3 \times 10^{2} {k J mol}^{- 1}$ . $1 g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298 K$ and $1$ atmospheric pressure. If heat capacity of the bomb calorimeter is $12.5 {k J K}^{- 1}$ , then final temperature of the system at the end of reaction is $x K$ . The value of $\frac{x}{10}$ is

30.00

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30.0

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Solution

Heat released by combustion = Heat taken by calorimeter.
Enthalpy of combustion of $1 g$ of graphite
$= \frac{300 k J}{12} = 25 k J$

$- 25 k J = - C Δ T$

$C = 12.5 k J$
$Δ T = 2 K$
$\Rightarrow T_{f} = 298 + 2$
$\Rightarrow T_{f} (x) = 300 K$
$∴ \frac{x}{10} = \frac{300}{10} = 30$

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Heat released by combustion = Heat taken by calorimeter. Enthalpy of combustion of 1 g of graphite =300 k J12=25 kJ −25 k J=−CΔT C=12.5 k J ΔT=2 K ⇒Tf=298+2 ⇒Tf(x)=300 K ∴x10=30010=30

Heat released by combustion = Heat taken by calorimeter.
Enthalpy of combustion of $1 g$ of graphite
$= \frac{300 k J}{12} = 25 k J$

$- 25 k J = - C Δ T$

$C = 12.5 k J$
$Δ T = 2 K$
$\Rightarrow T_{f} = 298 + 2$
$\Rightarrow T_{f} (x) = 300 K$
$∴ \frac{x}{10} = \frac{300}{10} = 30$