`C(s)("diamond") rarr C(g), DeltaH =- 714.8 kJ`
`DeltaH` is the heat of …………….of diamond.

C(s)("graphite") rarr C(g), DeltaH = 716.7 kJ DeltaH is the heat of ……………….of graphite.

H_(2)O(l) rarr H_(2)O(g), DeltaH = +40.7 kJ DeltaH is the heat of ……………..of water.

H_(2)O(g) rarr H_(2)O(l), DeltaH =- 40.7 kJ DeltaH is the heat of……………….of water.

H_(2)O(l) rarr H_(2)O(s), DeltaH =- 6.01 kJ DeltaH is the heat of ……………….of water.

H_(2)O(s) rarr H_(2)O(l), DeltaH = 6.01 kJ DeltaH is the heat of …………………of ice.

C_(4)H_(10)+(13)/(2)O_(2)(g)rarr4CO_(2)(g)+5H_(2)O(l), DeltaH =- 2878 kJ DeltaH is the heat of……….of butane gas.

Given that C_("(graphite)") rarr C_((g)), DeltaH=+716.7 kJ . C_("(diamond)") rarr C(g), DeltaH=714.8 kJ . The DeltaH for the following reaction is C_("(graphite)") rarr C_("(diamond)")

Given that: i. C("graphite") +O_(2)(g) rarr CO_(2)(g), DeltaH =- 393.7 kJ ii. C("diamond") rarr C("graphite"), DeltaH =- 2.1 kJ a. Calculate DeltaH for buring of diamond of CO_(2) . b. Calculate the quantity of graphite that must be burnt to evolve 5000 kJ of heta.

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