Home
Class 12
CHEMISTRY
Enthalpy change for the reaction, 4H(g...

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

A

`-869.9 KJ`

B

`+434.8 KJ`

C

`+217.4 KJ`

D

`-434.8` KJ

Text Solution

Verified by Experts

The correct Answer is:
B
`4H(g) to 2H_(2)(g), DeltaH=-869.6 KJ`
`2H_(2)(g) to 4H(g), DeltaH=869.6 KJ`
`H_(2)(g) to 2H(g)`
Dissociation energy of H-H bond
`=(869.6)/(2)=434.8 KJ `
Promotional Banner

Similar Questions

Explore conceptually related problems

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

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 :

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 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.

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)

Given that C(g)+4H(g)to CH_(4)(g) , Delta H =-166 kJ . The bond energy of the C - H bond will be

The dissociation energies of H_2and O_2 are 104 and 118 " kcal mol"^(-1) respectively. The heat of reaction 1/2H_2(g) +1/2O_2(g) rarrO-H(g) is 10 kcal. The bond energy of O - 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)

The enthalpy change (Delta H) for the reaction N_(2) (g)+3H_(2)(g) rarr 2NH_(3)(g) is -92.38 kJ at 298 K . What is Delta U at 298 K ?

Enthalpy change for reaction 1//2H_(2)(g)+1//2Cl_(2)(g)toHCl(g) is called :-