At 25°C, the following heats of reaction are known: $\Delta$H (kJ/mol) 2ClF + O₂ $\rightarrow$ Cl₂O + F₂O 167.4 2ClF₃ + 2O₂ $\rightarrow$Cl₂O + 3F₂O 341.4 2F₂+ O₂ $\rightarrow$2F₂O -43.4 At the same temperature, calculate $\Delta$H for the following reaction:

Question

Quizplus · Accepted Answer

To solve this problem, we need to use Hess's law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps. We can write the desired reaction as the sum of the first two reactions with a reverse of the third reaction: 11eaaf8d_c14a_1b0e_892c_d90a47594e90_TB6420_01 We can cancel out the ClF₂O and O₂ on both sides, and we get: F1F1F1S1 + 2F₂ --> F1F1F10 + 2F₂O Now we can use the given enthalpies of reaction to calculate the overall enthalpy change: F1F1F1S1 + 2ClF + O₂ --> F1F1F10 + Cl₂O + F₂O ΔH = 167.4 kJ/mol + 341.4 kJ/mol + (-1) x (-43.4 kJ/mol) = 552.2 kJ/mol F1F1F1S1 + 2F₂ --> F1F1F10 + 2F₂O ΔH = -552.2 kJ/mol (because the reaction is exothermic and ΔH is negative) Note that we reversed the sign of the enthalpy change for the third reaction because we reversed the reaction itself. Therefore, the correct answer is B, -108.7 kJ/mol.

At 25°C, the Following Heats of Reaction Are Known Δ\DeltaΔ H (KJ/mol) 2ClF + O2

At 25°C, the Following Heats of Reaction Are Known $\Delta$ H (KJ/mol)
2ClF + O₂