The enthalpy change for the reaction, `C_(2)H_(6)(g)rarr2C(g)+6H(g)` is X kJ. The bond energy of C-H bond is :

The enthalpy change for the reaction H_(2)(g)+C_(2)H_(4)(g)rarr C_(2)H_(6)(g) is ……………… The bond energies are, H - H=103, C-H=99, C-C=80 & C=C=145 "K cal mol"^(-1)

Enthalpy change for the reaction, 4H(g) to 2H_2(g) is -869.6 kJ The dissociation energy of H - H bond is

The enthalpy change for the reaction 2C("graphite")+3H_(2)(g)rarrC_(2)H_(6)(g) is called

Calculate the enthalpy change for the reaction CH_2 = CH_2(g) + H_2(g0 to CH_3 - CH_3(g) . The bond energies of C - C , C - H, H- H and C = C bonds are 83 kcal, 99 kcal 104 kcal and 145 kcal respectivley.

Enthalpy change for the reaction 2N_((g))+6H_((g))to2NH_(3(g))is-900 KJ The dissociation energy of N - H bond is

Calculate the enthalpy change for the reaction H_2(g) + 1/2 O_2(g) to H_2O (g) if the bond energies of H - H,O = O and O - H bonds are 104, 118 and 11 kcal mol^(-1) respectivley.

Calculate the enthalpies change for the reaction, H_(2)(g)+Cl_(2)(g)to2HCl(g) Given that, bond energies of H-H,Cl-Cl and H-Cl are 436, 243 and 432 kJ mol^(-1)

Determine enthalpy change for the reaction, CH_(4(g))+Cl_(2(g))rarr CH_(3)Cl_((g))+HCl_((g)) Bond energies for C-H,C-Cl,Cl-Cl,H-Cl are 412,338,242,431k J mol^(-1) respectively.

The enthalpy of atomisation for the reaction CH_(4) (g) rarr C (g) + 4H (g) " is " 1665 kJ mol^(-1) . What is the bond energy of C - H bond ?

The enthalpy change (DeltaH) for the process, N_(2)H_(4)(g)to 2N(g)+4H(g) is is 1724 kJ mol^(-1) . If the bond energy of N-H bond in ammonia is 391 kJ mol^(-1) , what is the bond energy for N-N bond in N_(2)H_(4) ?