Determine the enthalpy for the following reaction, given $\Delta$ (NH₃(g)) = - 46.1 kJ/mol, $\Delta$ (NO(g)) = 90.3 kJ/mol, and $\Delta$ (H₂O(g)) = -241.8 kJ/mol. 4NH₃(g) 5O₂(g) 4NO(g) 6H₂O(g) $~~~~~~~~$$\Delta$H_rxn = ? kJ

Question

Quizplus · Accepted Answer

We can use Hess's Law to solve this problem by manipulating the given reactions to obtain the desired reaction: 1) 4 NH₃(g) + F1F1F1S1F1F1F10F1F1F1S1F1F1F10 -> 4 H₂O(g) + F1F1F1S1F1F1F105O₂(g) 2) 5/2 NO(g) + F1F1F1S1F1F1F10F1F1F1S1F1F1F10 -> 5/2 N2(g) + F1F1F1S1F1F1F104NO(g) 3) H₂O(g) + F1F1F1S1F1F1F10H_rxn -> H2O(g) + SO2(g) + F1F1F1S1F1F1F109H₂O(g) Adding these reactions together gives: 4NH₃(g) + 5/2 NO(g) + H₂O(g) + F1F1F1S1F1F1F10F1F1F1S1F1F1F10 + F1F1F1S1F1F1F10H_rxn -> 2H2O(g) + SO2(g) + 4N2(g) + F1F1F1S1F1F1F10105.2 kJ/mol Using the enthalpies given in the problem, we can calculate the enthalpy of the desired reaction: Enthalpy = (4 x -46.1 kJ/mol) + (5/2 x 90.3 kJ/mol) + (-241.8 kJ/mol) + (-255.1 kJ/mol) = -905.2 kJ/mol Therefore, the answer is (D) (F1F1F1S1F1F1F10905.2 kJ/mol).

Determine the Enthalpy for the Following Reaction, Given Δ\DeltaΔ (NH3(g)) = - 46

Determine the Enthalpy for the Following Reaction, Given $\Delta$ (NH₃(g)) = - 46