At 25°C, the following heats of reaction are known: At the same temperature, use the above data to calculate the heat released (kJ)when 3.40 moles of ClF(g)reacts with excess F₂: ClF(g)+ F₂(g)$\rarr$ ClF₃(g) A)109 kJ B)233 kJ C)370.kJ D)465 kJ E)1,580 kJ

Question

Quizplus · Accepted Answer

The balanced chemical equation is: ClF(g) + F₂(g) → ClF₃(g) + F1F1F1S1 The overall ΔH° for the reaction is: ΔH° = ΣΔH°(products) - ΣΔH°(reactants) ΔH° = [(-328.5 kJ/mol) + (-79.9 kJ/mol)] - [(-247.3 kJ/mol)] ΔH° = -160.1 kJ/mol The heat released for 3.40 moles of ClF is: q = nΔH° q = (3.40 mol)(-160.1 kJ/mol) q = -544.34 kJ The negative sign indicates that the reaction releases heat, so we need to change the sign to get the final answer: q = -(544.34 kJ) ≈ -370 kJ Therefore, the best answer is (C), which is closest to our calculated value of -370 kJ.

At 25°C, the Following Heats of Reaction Are Known →\rarr→ ClF3(g) A)109 KJ B)233 KJ C)370

At 25°C, the Following Heats of Reaction Are Known $\rarr$ ClF₃(g)
A)109 KJ
B)233 KJ
C)370