Please help with this question...cant figure it out. Determine delta_r H^degree
ID: 1060398 • Letter: P
Question
Please help with this question...cant figure it out.
Determine delta_r H^degree for the following reaction, 2 NH_3(g) + 5/2 O_2(g) rightarrow 2 NO(g) + 3H_2O(g) given the thermochemical equations below. N_2(g) + O_2(g) rightarrow 2 NO(g) n_2(g) + 3H_2(g) rightarrow 2 NH_3(g) 2 NH_2 O(g) 2 H_2(g) + O_2(g) rightarrow 2 H_2 O(g) delta_r H degree = + 180.8 Kj/mol-rxn delta_r H degree = -91.8 Kj/mol-rxn delta_rH degree = 483.6 kJ/mol-rxn -1178.2 kJ/mol -452.8 kJ/mol -394.6 kJ/mol-rxn -211.0 kJ/mol-rxn + 1178.2 kJ/mol-rxnExplanation / Answer
N2(g) + O2(g) -----------> 2 NO(g) , drH0 = +180.8 kJ/mol-rxn - reaction 1
N2(g) + 3 H2(g) ----------> 2 NH3(g) , drH0 = -91.8 kJ/mol-rxn - reaction 2
2 H2(g) + O2(g) ----------> 2 H2O(g) , drH0 = -483.6 kJ/mol-rxn - reaction 3
2 NH3(g) + 5/2 O2(g) ----> 2 NO(g) + 3 H2O(g) , drH0 = ? - reaction 4
Rearrange reactions 1, 2 and 3 in such a way to get their sum equal to reaction 4. Reaction 4 is obtained by adding (reverse of reaction 2) + [Reaction 3 multiplied by (3/2)] + (reaction 1). The enthalpies are multiplied by the same factor (reaction 3) or reverse signed (reaction 2) or left undisturbed as to their respective reactions.
2 NH3(g) ----------> N2(g) + 3 H2(g) - Reverse of reaction 2
3 H2(g) + 3/2 O2(g) ----------> 3 H2O(g) - Reaction 3 multiplied by (3/2)
N2(g) + O2(g) -----------> 2 NO(g) - reaction 1
2 NH3(g) + 5/2 O2(g) ---------> 2 NO(g) + 3 H2O(g)
Now,
Following Hess’s law of constant heat summation-
drH0 (reaction 4) = (- drH0 of reaction 2) + (3/2 multiplied drH0 of reaction 3) + ( drH0 of reaction 1)
= (+ 91.8 kJ/mol) + (3/2) x (-483.6 kJ/mol) + (+180.8 kJ/mol)
= - 452.8 kJ/mol
Hence, drH0 of the reaction = - 452.8 kJ/mol-rxn