The decomposition of Ca(OH)2 into CaO(s) and H2O at constant pressure requires the addition of 109 kJ of heat per mole of Ca(OH)2 . b. Draw an enthalpy diagram for the reaction.
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Welcome back, everyone. Which of the following represents the enthalpy diagram for the reaction where magnesium two hydroxide solid decomposed to form magnesium oxide solid and vapor water and needed 243.3 kilojoules of heat per three moles of magnesium. Two hydroxide solid recall that the entropy change or heat content of a system is expressed by the term delta H. So again, this is the heat content of our system and our system would be our reaction in the case. And according to the prompt, we need, we'll say that the reaction needs 243.3 kilojoules of energy A K A heat per three moles of our only reactant being our magnesium to hydroxide. And we can actually use this ratio that we've written out to determine what our value for delta H is or our enthalpy change. And so in our calculators, we're gonna find a result equal to 81.1 kilojoules per mole. So this value that we found represents our amount of heat needed. I'm sorry. This says heat amount of heat needed for one mole of magnesium two hydroxide to form the products And to see the reaction, we have the decomposition of magnesium two hydroxide. This is going to decompose into magnesium oxide, solid and vapor water. Note that we have a positive value for our entropy positive 81.1 kilojoules per mole. And we should recall that a positive entropy value tells us that energy in the form of heat is absorbed by the system. So by our reaction, and we can say therefore, our products are going to be. And rather we can write it out this way, the energy of our products is going to be much more greater than the energy of our reactants. So we want to correspond our reaction to the diagram that places our products magnesium oxide and water at the top of the diagram representing a higher amount of energy. And so we can see that choices A and B both have magnesium oxide and water at the top of the diagram compared to where our reactant is located. However, in choice A, they calculated the entropy change to be 243.3 kilojoules for one mole of magnesium, two hydroxide to form our products. We know that that's not correct, it does not match our calculation here. And so, so far, we can say that B is definitely a correct answer choice because it represents specifically an endothermic reaction diagram. Let's look at the rest of our choices to see if that matches up with what we determined for the correct choice. So looking at choice c again, we have an incorrect calculation for entropy and also note that they place our reactant magnesium two hydroxide at higher energy than our products. This is going to actually represent an exothermic reaction diagram since energy is going to be released by our system in the form of heat given off, thereby decreasing the energy of our system and ultimately lowering the energy of our products more than it would lower our energy of our reactants. And so we have our reactant at higher energy than the products for an exothermic diagram. And lastly looking at choice D, we again have the reactant at higher energy than our products. And because we know that based on what the prompt tells us, our system being our reaction absorbs energy in the form of heat, we must have our products at a higher level of energy than our reactants. And so we know that choice D is incorrect. Although it does have the correct enthalpy value, we know it's still a incorrect choice because the energy of our products should be higher than our reactants for our endothermic diagram. And I just want to make another note that the B value for choice C which is an exothermic diagram should be negative if it were to be a correct answer. Choice. But otherwise our only correct answer choice is going to be choice B which is the only diagram that correctly displays our enthalpy of 81.1 kilojoules of heat that is absorbed, thereby increasing the energy of our system and placing our products, magnesium oxide and water at the higher energy level than our reactant. So I hope that this was helpful and please let us know if you have any questions.