10. Ammonia burns in oxygen according to the following equation: 4 NH3 (g) + 502

Question

10. Ammonia burns in oxygen according to the following equation: 4 NH3 (g) + 502(g) 4NO (g) +6 H20 (g) Calculate the enthalpy change (AH) for this reaction based on the following thermochemical equations: (10 points) N2(g)+O2 (g) 2NO (g) ??-180.6 kJ N2 (g) 3 H2 (g)2 NH3 (g) H -91.8 kJ 6 H2 (g) +302 (g) 6H20 (g) AH 967.4kJ Clearly show all manipulations which you perform to each reaction above in calculating the final answer! 11. What is the smallest atom in period three of the periodic chart? 12. Which ion below has a noble gas electron configuration? a)b) Kc) C d e) 13. For which element would the outer shell electrons be described as 3s3p a)Mg b)C 14. What element has the electron configuration 1s22822p63s23p64s23d10? c) Si d) S e) O a. Zn b. Ge ?.?? d. Ni 15. The transition metal manganese (Mn) has a specific heat of 0.480 J/g C, 60.00 grams of the manganese, at a temperature of 99.6?, is placed in a calorimeter containing water at 24.3°C. The final temperature of the water and metal is 29.3°C. The specific heat of water is 4.184 J/g C. a. what is the value of q for the cooling of the manganese metal? b. What was the mass of the water in the calorimeter? 16. In which list are all of the elements in the same period? a) K, Ca, Sr b) C, P, Se e) Ce, Th, U d) F. Cl, Br e) Sc, Cr, Ni 17. If you perform the reaction below, using 10.00 moles of Fe and 8.000 moles of O2 how much heat would be released (in kJ)? a. 1652 kJ b. 413 kJ c. 551 kJ d. 4130. kJ e. 4405 kJ

Explanation / Answer

4NH3 (g) + 5O2 (g) --------------> 4NO (g) + 6H2O (g) -----------1

N2 (g) + O2 (g) ------à 2NO (g)    ?H = 180.6 KJ ---------------2

N2 (g) + 3H2 (g) ------à 2NH3 (g)    ?H = -91.8 KJ -------------3

6H2 (g) + 3O2 (g) +------à 6H2O (g)    ?H = -967.4 KJ ------ -4

We have to multiply with factors and rearrange the three equations 2,3 and 4, in such a way that we get the overall reaction 1. And each time a reaction 2,3 and 4 is factored, the ?H of that reaction will be factored as well.

There are 4 moles of NO in parent eqyuation1 and only 2 moles of NO in equation 2, so one would multiply equation 2 with a factor 2.

N2 (g) + O2 (g) ------à 2NO (g)    ?H = 180.6 KJ ---------------2

2x (N2 (g) + O2 (g) ------à 2NO (g) )   ?H = 2 x180.6 KJ

2N2 (g) + 2O2 (g) ------à 4NO (g)    ?H = 361.2 KJ ---------------5

We have ammonia in equation 3, but only 2 moles are produced of NH3 are produced. But in the question, there are 4 moles, so one would multiply equation 3 by factor 2 and since NH3 is on the reactants on the overall equation, we would reverse the equation.

2 x (N2 (g) + 3H2 (g) ------> 2 x 2NH3 (g) =====)    ?H = 2* -91.8 KJ

2 N2 (g) + 6H2 (g) ------> 4 NH3 (g) =====)    ?H = -183.6 KJ

Reversing the reaction, ?H of the reaction also reverses in its sign

4 NH3 (g) =====> 2 N2 (g) + 6H2 (g) ?H = +183.6 KJ ------   6

Now we put together the three equations 5,6 and 3 and cancel out common terms

2N2 (g) + 2O2 (g) ------> 4NO (g)    ?H = 361.2 KJ ---------------5

4 NH3 (g) =====> 2 N2 (g) + 6H2 (g) ?H = +183.6 KJ ------   6

6H2 (g) + 3O2 (g) +------> 6H2O (g)    ?H = -967.4 KJ ------ -4

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4NH3 (g) + 5O2 (g) --------------> 4NO (g) + 6H2O (g) ?H = 361.2 + 183.6 – 967.4 = -422.6 KJ

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