1 g of graphite is burnt in a bomb calorimet | vedclass

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Question

(B) The heat absorbed by the calorimeter is given by $q = C_v 	imes \Delta T$.
Given $C_v = 20.7 \ kJ \ K^{-1}$ and $\Delta T = (299 - 298) \ K = 1

Vedclass · Accepted Answer

$-248.4 \ kJ \ mol^{-1}$

Vedclass · Answer

(B) The heat absorbed by the calorimeter is given by $q = C_v 	imes \Delta T$.
Given $C_v = 20.7 \ kJ \ K^{-1}$ and $\Delta T = (299 - 298) \ K = 1 \ K$.
Heat released by the reaction $(q_{rxn})$ is equal to $-q_{calorimeter} = -20.7 \ kJ \ K^{-1} 	imes 1 \ K = -20.7 \ kJ$.
This is the internal energy change $(\Delta U)$ for the combustion of $1 \ g$ of graphite.
For $1 \ mol$ of graphite $(12 \ g)$,the internal energy change $(\Delta U)$ is:
$\Delta U = \frac{-20.7 \ kJ}{1 \ g} 	imes 12 \ g \ mol^{-1} = -248.4 \ kJ \ mol^{-1}$.
Since the reaction is $C \ (graphite) + O_{2(g)} ightarrow CO_{2(g)}$,the change in moles of gaseous species is $\Delta n_g = 1 - 1 = 0$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$,since $\Delta n_g = 0$,we have $\Delta H = \Delta U$.
Therefore,$\Delta H = -248.4 \ kJ \ mol^{-1}$.

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