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Q.

One gram of graphite is burnt in a bomb calorimeter in excess of dioxygen at 298K and 1 atm pressure. During this reaction temperature increased from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 KJ/K the enthalpy change of the reaction $C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g)$ is

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a

– 248 KJ / mol

b

+ 248 KJ / mol

c

– 20.7 KJ / mol

d

+ 20.7 KJ / mol

answer is A.

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Detailed Solution

$q = - C_{V} . Δ T = - 20.7 \times (299 - 298) = - 20.7 K J$
(Here, - ve sign indicates the exothermic nature of combustion reaction)
$Δ U = - 20.7 \times 12 = - 248 K J / m o l e$
$\begin{array}{l} Δ H = Δ U (a s Δ n (g) = 0) \\ \Rightarrow Δ H = - 248 K J / m o l e \end{array}$

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One gram of graphite is burnt in a bomb calorimeter in excess of dioxygen at 298K and 1 atm pressure. During this reaction temperature increased from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 KJ/K the enthalpy change of the reaction C(graphite)+O2(g)→CO2(g) is