The following reactant molecules are rearranged to form the product molecules shown. The relevant bond enthalpies are given, in kJ/mol. What is the estimated enthalpy change for this reaction?

ReactionEnergy

A. 678 kJ/mol

B. 198 kJ/mol

C. –265 kJ/mol

D. –632 kJ/mol

Correct B. 198 kJ/mol

Explanation:

In this reaction, four C–H bonds and two O–H bonds are broken, so the total energy absorbed is 4 · 413 + 2 · 463 = 1,652 + 926 = 2,578 kJ/mol. For the products, one C=O bond and three H–H bonds are formed, so the energy released is 1,072 + 3 · 436 = 1,072 + 1,308 = 2,380 kJ/mol. The total enthalpy change is equal to the energy absorbed minus the energy released, so we have 2,578 – 2,380 = 198 kJ/mol. The enthalpy change is positive, so the reaction is endothermic.

the ratio of the conjugate base to the acid required to attain the desired ph.

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.