7) Use the following thermochemical equations C 2 H 2 (g) + 5/2 O 2 (g) 2 C O 2

Question

7) Use the following thermochemical equations C 2 H 2 (g) + 5/2 O 2 (g) 2 C O 2 (g) + H 2 O (g) H ° = - 1300 k J C 2 H 6 (g) + 7/2 O 2 (g) 2 CO 2 (g) + 3 H 2 O (g) H ° = - 1560 k J H 2 (g) + 1/2 O 2 (g) H 2 O (l) H ° = - 286 k J to calculate H ° for the following reaction: C 2 H 2 (g) + 2 H 2 (g) C 2 H 6 (g) 7) 8) Given the thermochemical equation: 2 HI (g) H 2 (g) + I 2 (s) H ° = - 52.96 k J , what is the heat of formation of HI (g), in k J /mol? 8) 9) How much heat is absorbed/released when 20.00 g of NH 3 ( g ) reacts in the presence of excess O 2 ( g ) to produce NO( g ) and H 2 O( l ) according to the following chemical equation? 4 NH 3 ( g ) + 5 O 2 ( g ) 4 NO( g ) + 6 H 2 O( l ) H ° = + 1168 kJ

Explanation / Answer

C 2 H 2 (g) + 5/2 O 2 (g) -------> 2 C O 2 (g) + H 2 O (g) H ° = - 1300 k J

C 2 H 6 (g) + 7/2 O 2 (g) -----------> 2 CO 2 (g) + 3 H 2 O (g) H ° = - 1560 k J

H 2 (g) + 1/2 O 2 (g) ------------> H 2 O (l) H ° = - 286 k J

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C 2 H 2 (g) + 2 H 2 (g) -----------------> C 2 H 6 (g) ; H ° = -312 kJ

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2 HI (g) ----------------> H 2 (g) + I 2 (s) H ° = - 52.96 k J

deltaHrxn = deltaHf (products) - deltaHf (reactants)

deltaHrxn = 0 - ( 2 x H-I )

=> -52.96 = -2 x H-I

=> heat of formation of H-I = 26.48 kJ

4 NH 3 ( g ) + 5 O 2 ( g ) 4 NO( g ) + 6 H 2 O( l ) H ° = + 1168 kJ

4 moles of NH3 = 4 x 17 = 68 g

68 g NH3 reaction with excess oxygen needs 1168 kJ of energy.

Energy absorbed when 20 g of NH3 reacted with excess oxygen = 20 x 1168 / 68 = 343.53 kJ

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