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The enthalpies of combustion ofC(graphite) and C( diamond) are 393.5 kJ mol1 and 395.3 kJ mol1, respectively. The enthalpy change of the reaction C(graphite) __, C(diamond) is

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a
1.80 kJ mol1
b
1.80 kJ mol1
c
3.60 kJ mol1
d
3.60 kJ mol1

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detailed solution

Correct option is B

 (a) C (graphite )+O2(g)CO2(g) ΔHa=393.5 kJ mol1

 (b) C( diamond )+O2(g)CO2(g) ΔHb=395.3 kJ mol1

The given transformation C(graphite)  C(diamond) is obtained by subtracting Eq. (b) from Eq. (a). Hence

ΔH=ΔHbΔHa=[393.5(395.3)] kJ mol1=1.80 kJ mol1

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