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Calorimeter

constant pres...

Question

constant pressure.
The thermal capacity of calorimeter system is $17.7 k J$ $K^{- 1}$ . ( $R = 8.313 {m o l}^{- 1} K^{- 1}$ ) (only magnitude in nearest integer in kj/mol)

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Solution

$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ ; $T = 27^{o} C = 300 K$
At constant pressure
$Δ H = Δ U + Δ n R T$
Rise in temperature of the system = ${0.5}^{o} C = 0.5 K$
Heat released to the system $= - C_{p} Δ T = - 17.7 \frac{k J}{K} \times 0.5 K = - 8.85 k J$
0.16 g of methane $= \frac{0.16}{16}$ mole $= 0.01$ mole of methane (molecular weight of methane =16 g)
$Δ U = \frac{- 8.85}{0.01} = - 885 k J m o l e^{- 1}$
$Δ n =$ moles of gaseous products $-$ moles of gaseous reactants
$Δ n = 1 - (1 + 2) = - 2$
$R = 8.313 J m o l e^{- 1} K^{- 1} = 8.313 \times 10^{- 3} k J m o l e^{- 1} K^{- 1}$
$Δ H = Δ U + Δ n R T$
$\Rightarrow Δ H = - 885 k J m o l e^{- 1} + (- 2 m o l e) \times 8.313 \times 10^{- 3} k J m o l e^{- 1} K^{- 1} \times 300 K = - 889.98 k J m o l e^{- 1}$
Answer = 890

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