1) The chemical reaction is as follow Mg(s) + N2(g) + 3O2(g) ---> Mg(NO3)2(s) 2)

Question

1) The chemical reaction is as follow

Mg(s) + N2(g) + 3O2(g) ---> Mg(NO3)2(s)

2) rearrange the three data equations to yield the above reaction when added together, the first equation must be reversed so that the Mg(NO3)2 is on the right-hand side.

Mg3N2(s) + 6MgO(s) ---> 8Mg(s) + Mg(NO3)2(s) H° = +3280.88

3) Notice the 8Mg. We have to make that go away while leaving one Mg on the left- hand side.

3Mg(s) + N2(g) ---> Mg3N2(s) H° -461.08

2Mg(s) + O2(g) ---> 2MgO(s) H° -1203.60

4) Multiply the third equation by 3:

6Mg(s) + 3O2(g) ---> 6MgO(s) H° -3610.80

5) It gives us 9Mg from the 3Mg in the second data equation and 6Mg in the third. Finally we have the following data equations:

Mg3N2(s) + 6MgO(s) ---> 8Mg(s) + Mg(NO3)2(s) H° = +3280.88

3Mg(s) + N2(g) ---> Mg3N2(s) H° = -461.08

6Mg(s) + 3O2(g) ---> 6MgO(s) H° = -3610.80

6) Now we need to add all, the Mg3N2, the 6MgO and 8Mg will cancel. Adding the three enthalpies for the final answer:

+3280.88 + (-461.08) + (-3610.80) = -791 kJ

Explanation / Answer

22) Calculate the ', for Mg(NO3)2(s) from the following data. 8 Mg(s) + Mg(NOJ2(s) -> Mgth(s) + 6 MgO(s) H.--3280.88 2 MgO(s) 2 Mg(s) + O2(g) h"-+1203.60 The chemical reaction for the AH' for Mg(NOs)2(s) is this: Mg(s) + N2(g) + 3 O2(g) Mg(NO3)2(s) Ans: -791 kJ) 23) Given the following thermochemical equations 2H2(g)+O2(g) N205(g) + H2O() 2H20(1) 2 HNO3 HP =-571.6 kJ HP-73.7 kJ Ho--1 74.1 kJ Calculate H for the formation of one mole of dinitrogen pentoxide from its elements in their stable state at 25 C and 1 atm. N2(g) + O2(g)--> N205(g) (Ans: +11.3 kJ)

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