Question #162466

a.)

N_{2}H_{4} and H_{2}O_{2 }react according to the equation below.

N_{2}H_{4}(g)+2H_{2}O_{2}(g)--> N_{2}(g)+ 4H_{2}O(g)

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

Delta H*/KJmol^{-1} N_{2}H_{4}(g)= +75 H_{2}O_{2}(g)=-133 H_{2}O(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

Expert's answer

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

N_{2}H_{4}= 1N-N + 4N-H = 163 + 4(388) = 1715KJ

2H_{2}O_{2}= 2[2H-O + 1O-O]= 2[2(463) + 1(x)] = 2x + 1852

N_{2}= 1N-N = 944KJ

4H_{2}O= 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

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