Q1Calculate the delta H for N2 (g) + O2 (g)------>2NO(g) 4NH3(g) +5O2(g)--->4NO(

Question

Q1Calculate the delta H for N2 (g) + O2 (g)------>2NO(g)
4NH3(g) +5O2(g)--->4NO(g)+6H2O(l). DeltaH= -1170kJ 4NH3(g) + 3O2(g)--->2N2(g)+6H2O(l) Delta H =-1530kJ
Q2Calculate standard heat of formation delta H degree f. For NCl 3
NH3(g)+3HCl (g)--->NCl3(g)+3H2(g). Delta H =564.8kJ
Q1Calculate the delta H for N2 (g) + O2 (g)------>2NO(g)
4NH3(g) +5O2(g)--->4NO(g)+6H2O(l). DeltaH= -1170kJ 4NH3(g) + 3O2(g)--->2N2(g)+6H2O(l) Delta H =-1530kJ
Q2Calculate standard heat of formation delta H degree f. For NCl 3
NH3(g)+3HCl (g)--->NCl3(g)+3H2(g). Delta H =564.8kJ

4NH3(g) +5O2(g)--->4NO(g)+6H2O(l). DeltaH= -1170kJ 4NH3(g) + 3O2(g)--->2N2(g)+6H2O(l) Delta H =-1530kJ
Q2Calculate standard heat of formation delta H degree f. For NCl 3
NH3(g)+3HCl (g)--->NCl3(g)+3H2(g). Delta H =564.8kJ

Explanation / Answer

consider the given reactions

1) 4 NH3 + 5O2 ---> 4 NO + 6 H20 dH1 = -1170

2) 4NH3 + 3 O2 ---> 2N2 + 6H20 dH2 = -1530

3) N2 + 02 ---> 2NO

we can see that

eq3 = ( eq1 - eq2 ) / 2

so

dH3 = ( dH1 - dH2) / 2

dH3 = ( -1170 + 1530) / 2

dH3 = 180

so

dH for N2 + 02 ---> 2 NO is 180 kJ

2)

consider the given reaction

NH3 + 3 HCl --> NCl3 + 3 H2

we know that

dHrxn = dHfo products - dHfo reactants

so

dHrxn = (dHfo NCl3) + ( 3 x dHfo H2 ) - (dHfo NH3) - ( 3 x dHfo HCl)

564.8 = dHfo NCl3 + ( 3 x 0) - ( -45.9) - ( 3 x -92.3)

dHfo NCl3 = 242

so

standard heat of formation of NCl3 is 242 kJ / mol

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