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katrin [286]
3 years ago
6

Consider the reaction between hydrazine and hydrogen to produce ammonia, N2H4(g)+H2(g)→2NH3(g). Use enthalpies of formation and

bond enthalpies to estimate the enthalpy of the nitrogen-nitrogen bond in N2H4. Express your answer in kilojoules to 3 significant figures.
Chemistry
1 answer:
sammy [17]3 years ago
6 0

Answer:

The answer is explained below:

Explanation:

Given the chemical equation:

                   N2H4(g)+H2(g)→2NH3(g)

The standard enthalpy of formation is given by the following formula:

            ΔH^0 rxn = ∑ B reactants - ∑ product

       -187.78 kJ/mol = [( 1x B N-N) + (1 x B N-H ) + (1 x b H-H)] - [ 6 x B N-H]

       -187.78 kJ/mol = [( 1x B N-N)  +(4 x 391 kJ/mol) + (1 x 436  kJ/mol)] - [ 6x 391  kJ/mol ]

        -187.78 kJ/mol =  B N-N  + (1564 + 436 - 2346) kJ/mol

                                       B   = 158.22 kJ/mol

So, in this case the enthalpy of N-N bond is 158 kJ/mol

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Which of the following is true of solids?
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            BaCrO4(s)    →  Ba^2+(aq)    +   CrO4^2-(aq)

initial                               0                          0

change                          +X                       +X 

Equ                                  X                         X


when Ksp = [Ba^2+][CrO4^2-]

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According to the reaction equation:

            BaCrO4(s)  →  Ba^2+(aq)    +   CrO4^2-(aq)

initial                                 0                      0.0016

Change                           +X                      +X

Equ                                   X                      X+0.0016

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by substitution:

2.1 x 10^-10 = X*(X+0.0016) by solving for X 

∴ X = 1.3 x 10^-7

∴ solubility =X = 1.3 x 10^-7

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3 years ago
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