K(s) + 1/2 Cl2(g) → KCl(s) ΔH = -437 kJ/mol The elements K and Cl react directly to form the compound KCl according to the equation above. Refer to the information above and the table to answer the questions that follow.

Cl2(g) + 2e- → 2Cl-(g)

Which of the following expressions is equivalent to ΔHo for the reaction represented above?

Process ΔH° (kj/mol)
K(s) → K(g) v
K(g) → K+(g) + e^- w
Cl2(g) → 2Cl(g) x
Cl(g)+ e^- → Cl^-(g) y
K+(g) + Cl(g)^- → KCL z

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.