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Question

Calculate the enthalpy change accompaning the conversion of 10 g of graphite into diamond if the heats of combustion of C(graphite) and C(diamond) are 94.05 kcal and 94.50 kcal respectively.

A
4.5 kcal
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B
0.45 kcal
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C
0.375 kcal
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D
3.75 kcal
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Solution

The correct option is C 0.375 kcal

Given that:
(i) C(graphite)+ O2(g) CO2(g) ΔH1=94.05 kJ mol1

(ii) C(diamond)+ O2(g) CO2(g) ΔH2=94.50 kJ mol1

now, [eqn. (i) - eqn. (ii)] we get :

C(graphite)+ O2(g)C(diamond) O2(g) CO2(g) CO2(g)
For this reaction ΔH=ΔH1ΔH2

Simplifying the above equation we get,

C(graphite) C(diamond)
Also putting values of ΔH1and ΔH2
ΔH= 94.05(94.50)=0.45 kcal

Since this enthalpy change is only for conversion of 1 mole, i.e. ,12 g of C(graphite) to C(diamond), therefore, for the conversion of 10 g of C(graphite) to C(diamond)

ΔH=0.45×1012=0.375 kcal

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