At `25^(@)C,DeltaH_(f)^(0)=`0 for graphite, but for diamond `DeltaH_(f)^(0)ne0`. Why?

H_(2)(g)+(1)/(2)O_(2)(g) to H^(+)(aq)+OH^(-)(aq),DeltaH^(0)=-228.5kJ*mol^(-1) . If DeltaH_(f)^(0)[H^(+)]=0 , then what will be the value of DeltaH_(f)^(0)[OH^(-)] ?

The formation of the oxide ion O^(2-)(g) , form oxygen atom requires first an exothermic and then an endothermic step as shown below, O(g)+e^(-) to O^(-)(g),DeltaH_(f)^(0)=-141kJ*mol^(-1) O^(-)(g)+e^(-) to O^(2-)(g),DeltaH_(f)^(0)=+780kJ*mol^(-1) ltbr. Thus, process of formations of O^(2-) in gas phase is unfavourable even through O^(2-) is isoelectronic with neon. it is due to the fact that-

Between O_(2)(g) and O_(3)(g) , whose standard enthalpy of formation (DeltaH_(f)^(0)) is taken as zero at 1 atm and 25^(@)C ?

H_(2)O(l)toH^(+)(aq)+OH^(-)(aq),DeltaH^(0)=57.32kJ*mol^(-1) H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH^(0)=-285.8kJ*mol^(-1) at 25^(@)C . If DeltaH_(f)^(0)[H^(+)(aq)]=0 , then the standard heat of formation (kJ*mol^(-1)) for OH^(-)(aq) at 25^(@)C is-

Given: C(graphite, s) +O_(2)(g)toCO_(2)(g),DeltaH^(0)=x_(1) C(graphite, s) +(1)/(2)O_(2)(g) to CO(g),DeltaH^(0)=x_(2) CO(g)+(1)/(2)O_(2)(g) to CO_(2)(g),DeltaH^(0)=x_(3) Express x_(3) in terms of x_(1) and x_(2) .

Some reaction and their DeltaH^(0) values are given below: C(graphite, s) +(1)/(2)O_(2)(g)toCO(g),DeltaH^(0)=a kJ CO(g)+(1)/(2)O_(2)(g)toCO_(2)(g),DeltaH^(0)=b kJ H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH^(0)=c kJ CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l),DeltaH^(0)=d kJ 2C(gaphite, s) +3H_(2)(g)toC_(2)H_(6)(g),DeltaH^(0)=m kJ Which of the following statements are correct-

In which of the following reactions does DeltaH^(0) at 25^(@)C indicate the standard enthalpy of formation (DeltaH_(f)^(0)) of the compound formed in each of the reactions? (2) N_(2)(g)+3H_(2)(g)to2NH_(3)(g) (2) (1)/(2)N_(2)(g)+(1)/(3)O_(3)(g)toNO(g) .

Given: NH_(3)(g)+3Cl_(2)(g)toNCl_(3) (g)+3HCl(g),-DeltaH_(1) N_(2)(g)+3H_(2)(g)to2NH_(3)(g),DeltaH_(2) H_(2)(g)+Cl_(2)(g)to2HCl(g),DeltaH_(3) Heat of formation (DeltaH_(f)) of NCl_(3) in terms of DeltaH_(1), DeltaH_(2) and DeltaH_(3) is-

Determine the standard enthalpy of formation of isoprene(g) at 298K temperature. Given: at 298K, DeltaH^(0)(C-H)=413kJ*mol^(-1) DeltaH^(0)(H-H)=436kJ*mol^(-1) DeltaH^(0)(C-C)=346kJ*mol^(-1) , DeltaH^(0)(C=C)=611kJ*mol^(-1) , C("graphite, s")toC("graphite, g"), DeltaH^(0)=717kJ*mol^(-1)