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Question

If the enthalpy of combustion of diamond and graphite are 395.4kJmol1 and 393.6kJmol1,
The enthalpy change for the C(graphite)C(diamond) is :

A
ΔH=1.8kJmol1
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B
ΔH=2.8kJmol1
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C
ΔH=5.8kJmol1
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D
ΔH=6.8kJmol1
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Solution

The correct option is A ΔH=1.8kJmol1
C(diamond)+O2(g)CO2(g) .....ΔH=395.4kJmol1...(i)
C(graphite)+O2(g)CO2(g) ΔH=393.6kJmol1...(ii)

Subtracting equation (i) from equation (ii), we get-

C(graphite)C(diamond)

ΔH=393.6kJ(395.4kJ)

ΔH=+1.8kJmol1.

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