Answer to Question #162474 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_{f -} H_i.

. = -244 - (+75- 133)

= -186KJmol^-1

b) Mean bond enthalpy = (388 +163 + 944 +463 )KJmol^-1./ 4

= 489.5KJmol^-1