1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation C graphite- O 2 g- CO 2 g During the reaction- temperature rises from 298 K to 299 K- If the heat capacity of the bomb calorimeter is 20-7 kJ-K- what is the enthalpy change for the above reaction at 298 K and 1 atm-

C+O_2→ CO_2 Δ ng=0 Δ H=ov+Δ ngRT Δ v=qv= cv×Δ T q_v= 20.7× 10^3× 0.8 = 1705.6 J per 0.562g amount of heat liberated during on ...

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Standard XII

Chemistry

Dipole Moment

1 g of graphi...

Question

1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation
$C (g r a p h i t e) + O_{2} (g) \to {C O}_{2} (g)$
During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?

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Solution

$C + O_{2} \to C O_{2}$
$Δ n g = 0$
$Δ H = o v + Δ n g R T$
$Δ v = q v = - c v \times Δ T$
$q_{v} = - 20.7 \times 10^{3} \times 0.8$
$= - 1705.6$ J per $0.562 g$
amount of heat liberated during one mole carbon
$= \frac{1705.6 \times 12}{0.562} = 36 \times 20.2 J$
$Δ H = Δ ν + Δ n g R T$
so, $Δ H = 00$
So , $Δ H = 36420.2 J$

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C+O2→CO2Δng=0ΔH=ov+ΔngRTΔv=qv=−cv×ΔTqv=−20.7×103×0.8=−1705.6J per 0.562gamount of heat liberated during one mole carbon=1705.6×120.562=36×20.2JΔH=Δν+ΔngRTso, ΔH=00So ,ΔH=36420.2J