Home
Class 12
CHEMISTRY
Bond energies of H - H bond is 80 kJ/mol...

Bond energies of H - H bond is 80 kJ/mol, I - I bond is 100 kJ/mol and for H - I bond is 200 kJ/mol, the enthalpy of the reaction : `H_(2)(g) + I_(2)(g) rarr 2HI(g)` is

A

`-120` kJ

B

`200` kJ

C

`+100` kJ

D

`+120` kJ

Text Solution

Verified by Experts

The correct Answer is:
B
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • THERMODYNAMICS

    AAKASH INSTITUTE|Exercise SECTION-C|13 Videos

  • THERMODYNAMICS

    AAKASH INSTITUTE|Exercise SECTION-D|9 Videos

  • THERMODYNAMICS

    AAKASH INSTITUTE|Exercise SECTION-A|45 Videos

  • THE SOLID STATE

    AAKASH INSTITUTE|Exercise Assignment (SECTION - D) (ASSERTION-REASON TYPE QUESTION)|20 Videos

Similar Questions

Explore conceptually related problems

Bond energies of N-=N, H-H and N-H bonds are 945,463 & 391 kJ mol^(-1) respectively, the enthalpy of the following reactions is : N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)

The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction ? H_(2) (g) + Br_(2) (g) rarr 2HBr (g) . Given that, bond energy of H_(2), Br_(2) and HBr is 435 kJ mol^(-1), 192 kJ mol^(-1) and 368 kJ mol^(-1) respectively

Knowledge Check

  • Given the bond energies of N=N, H – H and N - H bonds are 945, 436 and 391 "kJ mol"^(-7) respectively, the enthalpy of the reaction N_(2(g)) + 3H_(2(g)) rarr 2 NH_(3(g)) is :

    A
    `-93 ` kJ
    B
    `102 ` kJ
    C
    `-93 ` kJ
    D
    105 kJ

  • If the bond energies of H-H, Br- Br and H -Br are 433, 192 and 364kJ mol^(-1) respectively, DeltaH^(@) for the reaction H_(2)(g) + Br_(2)(g) rarr 2HBr(g) is

    A
    `- 261kJ`
    B
    `+103kJ`
    C
    ` + 261kJ`
    D
    `-103kJ`

  • If the bond energies of H - H, Br - Br and HBr are 433, 192 and 364 kJ mol^(-1) respectively, the DeltaH^(@) for the reaction, H_(2)(g)+Br_(2)(g)rarr2HBr(g) is

    A
    `+ 261` kJ
    B
    `- 103` kJ
    C
    `- 261` kJ
    D
    `+ 103` kJ

    • Similar Questions

    Explore conceptually related problems

    The net enthalpy change of a reaction is the amount of energy required to break all the bonds an reactant molecules amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction : H_(2)(g) + Br_(2) rarr 2HBr(g) Given that bond energy of H_(2), Br_(2) and HBr is 435 kJ mol^(-1) , 192 kJ mol^(-1) and 368 kJ mol^(-1) respectively.

    The bond enthalpy of C - C, C = C, H - H and C - H bonds are 350,600,400 and 410 kJ · mol-1 respectively. The enthalpy change for reaction C_(2)H_(4(g)) +H_(2(g)) rarr C_(2)H_(6(g)) is

    The bond energies of C-C,C=C, H-H and C-H bonds are 350 kJ mol^-1 , 600 kJ mol^-1 , 400 kJ mol^-1 and 410 kJ mol^-1 respectively. The enthalpy of hydrogenation of ethane is _______________.

    The equilibrium constant for A(g)+B_(2)(g) hArr AB_(2)(g)" "k_(p)=100 at 522K . Structure of AB_(2) is like H_(2)O . If bond energy of A-B bond is 200 kJ//mol and that of B-B bond is 100 kJ//mol, the find Delta S^(@) of the above reaction :

    The bond energies of H - H and I - I bonds are 435 kJ mol^-1 and 150 kJ mol^-1 respectively. If DeltaH_f^o for HI is 26.5 kJ mol^-1 then bond enthalpy of H-I bond is

    Home
    Profile