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For the allotropic change represented by...

For the allotropic change represented by the equation C (graphit) `rarr` C (diamond), `Delta H = 1.9 kJ`. If 6g of diamond and 6 g of graphite are separately burnt to yield `CO_(2)`, the enthalpy liberated in first case is

A

Less than in the second case by 1.89kJ

B

More than in the second case by 1.89 kJ

C

Less than in the second case by 11.34 kJ

D

More than in the second case by 0.945 kJ

Text Solution

Verified by Experts

The correct Answer is:
D
C(diamond) `rarr` C(graphite), `Delta H = -1.89` kJ/mole
C(diamond) `+ O_(2) (g) rarr CO_(2)(g), Delta H= Delta H` ...(i)
C(graphite) `+O_(2)(g) rarr CO_(2)(g), Delta H = Delta H_(2)` ... (ii)
By Lavoisier and Laplace Law :
`CO_(2)(g) rarr ("graphite") + O_(2) (g), Delta H = -Delta H_(2)` ... (iii)
Adding equation (i) and (iii) using Hess' Law :
C (diamond) `rarr` C (graphite), `Delta H= Delta H_(1) - Delta H_(2)`
Thus `Delta H_(1)- Delta H_(2) = -1.89`kJ/mole
Both `Delta H_(1) and Delta H_(2)` are negative so the value of `Delta H_(1)` is more negative (more exothermic reaction) than that of `Delta H_(2)`
1 mole of C = 12g. of C
So the enthalpy difference per 6 g. of diamond and graphite `=(-1.89)/(2)= -0.945kJ`
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