For the change,`C_("diamond") rarr C_("graphite"), Delta H = -1.89 kJ` , if 6 g of diamond and 6 g of graphite are separately burnt to yield `CO_2` the heat liberated in first case is:

Options

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

Answer

D

Explanation

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`