Answer to Question #162466 in General Chemistry for Me

Question

 a.)

 N₂H₄ and H₂O₂ react according to the equation below.

 N₂H₄(g)+2H₂O₂(g)--> N₂(g)+ 4H₂O(g)

Using the information given in the table, determine the accepted enthalpy of reaction for the following reaction:

Delta H*/KJmol^-1  N₂H₄(g)= +75 H₂O₂(g)=-133 H₂O(g)=-242

b.)

some mean bond enthalpies are given below, based on the enthalpy of reaction calculated in part a, what is the mean bond enthalpy in kj/mol for the O-O bond?

Mean bond enthalpy in KJ/mol^-1 N-H=388 N-N=163 N=N=944 H-O=463

2 H--O--O--H --> N=N+4 H--O--H

Answer 1

a. ∆H= H_p - H_r

∆H= [4(-242)] - [75+2(-133)]

∆H= -968+191=-777KJ/mol

b. Let O-O bond energy be x

N₂H₄= 1N-N + 4N-H = 163 + 4(388) = 1715KJ

2H₂O₂= 2[2H-O + 1O-O]= 2[2(463) + 1(x)] = 2x + 1852

N₂= 1N-N = 944KJ

4H₂O= 4[2H-O]= 4[2(463)]= 3704KJ

Now,

∆H= H_p - H_R

-777= [944 + 3704] - [1715 + 1852 + 2x]

-777= 1081 - 2x

2x= 1081 + 777

2x= 1858

x= 1858/2 = 929KJ

Therefore,

Bond energy of O-O bond= 929KJ