The enthalpy change for chemical reaction is denoted as `DeltaH^(Theta)` and `DeltaH^(Theta) = H_(P)^(Theta) - H_(R)^(Theta)`. The relation between enthalpy and internal enegry is expressed by equation:
`DeltaH = DeltaU +DeltanRT`
where `DeltaU =` change in internal enegry `Deltan =` change in number of moles, `R =` gas constant.
For the change, `C_("diamond") rarr C_("graphite"), DeltaH =- 1.89 kJ`, if `6g` of diamond and `6g` of graphite are seperately burnt to yield `CO_(2)` the heat liberated in first case is

Options

A. Less than in the second case by 1.89 KJ
B. Less than in the second case by 11.34 KJ
C. Less than in the second case by 14.34 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 mol^(-1)`
For 6g conversion of diamond into graphite
`Delta H=(-1.89)/(2)=-0.945 KJ`
`(1)/(2)C_("Diamond")rarr (1)/(2)C_("Graphite"), Delta H=-0.945 KJ`