Question

3. (3 points) Hydrazine (N2H4) can react with oxyger in the following chemical reaction: NaH.(1 +0:0) - N2(g) + 2H20 (1) Calculate AH, for the reaction above, given the following data: 2NH3(g) + 3N20(g) - 4260) + 3H2000 AH,º=-1010. kJ/mol N20(g) + 3H2(g) → N2H (1) + H2O(1 AH-=-317 k!/inol 2NH3(g) + 40:09) - NzH4O + H20(1 AH,"=-143 kJ/mol H2(g) +4202(9) - H20(1) AH"=-286 kJ/mol

Answer 1

equation 2 multiply with 3 and subtract equation 1 and 2

2NH3(g) + 3N2O(g) ---------> 4N2(g) + 3H2O(l)               $\Delta$H⁰r = -1010KJ/mole

3N2O(g) + 9H2(g) --------> 3N2H4(l) + 3H2O(l)               $\Delta$H⁰r   = -951KJ/mole

(-)             (-)                       (-)               (-)                                        (+)

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2NH3(g) +3N2H4---------> 4N2(g) + 9H2(g)                       $\Delta$H⁰r = -59KJ/mole --------->5

equation 4 multiply with 9 and addition equation 4 and 5

2NH3(g) +3N2H4---------> 4N2(g) + 9H2(g)                       $\Delta$H⁰r = -59KJ/mole --------->5

9H2(g) + 9/2 O2(g) --------> 9H2O(l)                                  $\Delta$H⁰r   = -2574KJ/mole ----->4

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2NH3(g) + 3N2H4(l) +9/2O2(g) ---------> 4N2(g) + 9H2O(l)     $\Delta$H⁰r    = -2633KJ/mole -------> 6

subtract of 6 and 3 equations

2NH3(g) + 3N2H4(l) +9/2O2(g) ---------> 4N2(g) + 9H2O(l)     $\Delta$H⁰r    =- 2633KJ/mole -------> 6

2NH3(g) + 1/2O2(g) -------------------------> N2H4(l) + H2O(l)     $\Delta$H⁰r = -143KJ/mole

(-)              (-)                                             (-)             (-)                             (+)

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4N2H4(l) + 4O2(g) ----------> 4N2(g) + 8H2O(l)                       $\Delta$H⁰r = -2490KJ/mole

The equation divide by 4

N2H4(l) + O2(g) ----------> N2(g) + 2H2O(l)                       $\Delta$H⁰r = -622.5KJ/mole >>>>answer

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