<p>Hey,</p>
<p>Ok I completely understand the concept but my problem is applying it. Is there some type of procedure to follow that will guarantee the right answer. Like are there 2 rxns I should add first before proceeding etc. Thx.</p>
<p>I don't understand what you're asking...</p>
<p>ok let me give u an example,</p>
<p>Calculate delta H for the reaction: NO+O --> NO2</p>
<p>2O3 --> 3O2 delta H= -427 kJ</p>
<p>O2--> 2O delta H= 495 kJ</p>
<p>NO+ O3--> NO2+O2 delta H= -199 kJ</p>
<p>Ok so is there some type of procedure to solving this problem or is it just working it out based on instinct.</p>
<p>i think your supposed to get stuff to cancel out</p>
<p>reverse first equation; change sign of delta H
reverse second equation; change sign of delta H
multiply third equation by 2; multiply delta H by 2</p>
<p>3O2 --> 2O3 delta H=427 kJ
2O --> O2 delta H=-495 kJ</p>
<h2>2NO + 2O3 --> 2NO2 + 2O2 delta H=-398 kJ</h2>
<p>2NO + 2O --> 2NO2 delta H=-466
divide by 2 to get
NO + O --> NO2 delta H=-233 kJ</p>
<p>is that right?</p>
<p>1) Identify the reaction that looks most like the total reaction you're trying to reach. In this case, it's the third reaction, where NO is a reactant and NO2 is a product--just like in the overall reaction. This helps you know how to treat the other steps.</p>
<p>2) In reaction from step 1, there's an O3 reactant we want to cancel out. So look for another reaction containing O3. In the first reaction, there are 2 moles of O3, but it's on the same side as the O3 in the third reaction. So reverse the first reaction so that the 2O3 is on the right side, and double the third reaction so the O3s will cancel out.</p>
<p>3) Reverse the second reaction to get the single oxygen atom on the reactant side, where it needs to be in the overall reaction.</p>
<p>4) Now rewrite the three equations with the changes you've made. Change the sign on the ΔH values of the reactions you reversed, and double the ΔH value of the reaction you doubled. Your equations should look like this:
2NO + 2O3 --> 2NO2 + 2O2 ΔH = -398 kJ
3O2 --> 2O3 ΔH = +427 kJ
2O --> O2 ΔH = -495 kJ</p>
<hr>
<p>2NO + 2O --> 2NO2 ΔH = -466 kJ</p>
<p>5) This is double the reaction you want, so divide everything by 2.
NO + O --> NO2 ΔH = -233 kJ</p>
<p>There is a method to it, but mostly it comes from practice. I've had a lot of practice with Hess's Law, so for me it's kind of instinctive. But there is definitely also a trial and error quality to solving these problems.</p>
<p>Hope this helps, and good luck!</p>
<p>I got the same answer as everyone else. You get three explanations! :)</p>
<p>NO + O --> NO2</p>
<p>1) I only want NO, O, and NO2. The rest I need to cancel out.</p>
<p>2) The third reaction is exactly what I want except for the O3 and O2, so I can't change reaction direction. Therefore to cancel out the O3, I need to flip the first reaction to get 203 as a reactant. Then I divide that by 2 to get plain O3. That cancels. The H of the first reaction becomes 213.5 kJ</p>
<p>3) Need to flip the second reaction also to put the O2 on the product side for cancellation. H of second reaction becomes -495 kJ.</p>
<p>4) We almost have it, except that the O needed in the final is 2O in the second reaction. So we divide by 2. H of the second reaction becomes -247.5 kJ.</p>
<p>5) Now the O2s can easily cancel. Reaction one has 3/2 O2, reaction two has 1/2 O2 as a product, and reaction three has 1 O2 as a product.</p>
<p>6) Add up 213.5 -247.5 - 199 = -233kJ</p>
<p>7) is that right?</p>
<p>ok thx much for clearing that up for me. I was initially confused about whether it was all trial and error but I guess its some trial and error and some analyzation.</p>
<p>-233 kJ is the correct answer.</p>