Init 4/23/2014 by Daniel R. Barnes The author regards this subject as honors-level material. It is not intended for a non-honors chemistry class. Hess’s law never appeared on the CST. It’s tricky stuff. It’s difficult to teach. I’m working on getting better at it, but it’s difficult to teach right now, so chill the fluorine out.

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l) To save space on scratch paper, I recommend referring to the equations given in the question by Roman numerals. I The first given equation would be “ I ”, Roman numeral one. The second given equation would be “ II ”, Roman numeral two. II The third given equation would be “ III ”, Roman numeral three. III Equation III is special. It is what I will call our “target equation,” since it is the one that we need to find out  H for.  H = ? kJ

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l) Throughout this worksheet, remember that the main idea in Hess’s law is that you can add chemical equations together. When you add the original equations to get your final equation, you add their  H values to get the  H for the final equation.

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l) I have erased all the notations in the white area because I don’t want you to get in the habit of writing on question sheets, since you’re not allowed to do that on tests. I’m going to put stuff in this blue area to represent what you should put on your scratch paper for a problem like this.

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l) Let’s do this.

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H III = ? kJ Declare your unknown. This is your mission statement. I II III

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H III = ? kJ Examine the target equation and ask yourself some questions... What needs to be on the left? Which of the original equations has those things in them? What needs to be on the right? Looks like there’s some extra junk we don’t need. Maybe if we wish really hard, it’ll just go away...

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H III = ? kJ GOOD NEWS! In this question, everything we need on the left of our target equation can be found on the left in the original equations. The same is true for the right. Therefore, to figure  H for our target equation, we should just try simply adding the original equations.

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H III = ? kJ I + II : Sn(s) + Cl 2 (g) + SnCl 2 (s) + Cl 2 (g)  SnCl 2 (s) + SnCl 4 (l) We’re just adding equation I to equation II, so everything on the left of both goes on the left of the new equation. Same thing for the right.

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H III = ? kJ I + II : Sn(s) + Cl 2 (g) + SnCl 2 (s) + Cl 2 (g)  SnCl 2 (s) + SnCl 4 (l) Remember the stuff we don’t need? Guess what? Since it’s on both sides of the arrow in the new equation, we can cancel them out!

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H III = ? kJ I + II : Sn(s) + Cl 2 (g) + SnCl 2 (s) + Cl 2 (g)  SnCl 2 (s) + SnCl 4 (l) We can combine them. Notice that Cl 2 (g) appears twice on the left of the arrow. Let’s clean up this equation. Time to re-write it!

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H III = ? kJ I + II : Sn(s) + Cl 2 (g) + SnCl 2 (s) + Cl 2 (g)  SnCl 2 (s) + SnCl 4 (l) The two separate Cl 2 (g)’s combine to form 2Cl 2 (g). Also, the SnCl 2 (s)’s disappear. I + II : Sn(s) + 2Cl 2 (g)  SnCl 4 (l)

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H III = ? kJ I + II : Sn(s) + Cl 2 (g) + SnCl 2 (s) + Cl 2 (g)  SnCl 2 (s) + SnCl 4 (l) Look familiar? I + II : Sn(s) + 2Cl 2 (g)  SnCl 4 (l) Equation  &  combine to make equation III.

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H III = ? kJ I + II : Sn(s) + Cl 2 (g) + SnCl 2 (s) + Cl 2 (g)  SnCl 2 (s) + SnCl 4 (l) I + II : Sn(s) + 2Cl 2 (g)  SnCl 4 (l) BIG IDEA: If equations I &  comine to give us equation ...  H   H   H ... You can add their  H’s to get the  H of equation  = ( kJ) + ( kJ)  H  = kJ = answer choice d.

Sn(s) + Cl 2 (g)  SnCl 2 (s)  H = kJ 1. Consider the following equations: SnCl 2 (s) + Cl 2 (g)  SnCl 4 (l)  H = kJ If the above is true, what is  H for the following equation? Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H III = ? kJ I + II : Sn(s) + Cl 2 (g) + SnCl 2 (s) + Cl 2 (g)  SnCl 2 (s) + SnCl 4 (l) I + II : Sn(s) + 2Cl 2 (g)  SnCl 4 (l)  H   H   H  = ( kJ) + ( kJ)  H  = kJ = answer choice d.

[phew]

3. Examine the two equations below. 2Al(s) + 3/2 O 2 (g)  Al 2 O 3 (s)  H = kJ 2Fe(s) + 3/2 O 2 (g)  Fe 2 O 3 (s)  H = kJ Assuming that the above information is true, what is  H for the following equation? 2Al(s) + Fe 2 O 3 (s)  2Fe(s) + Al 2 O 3 (s)  H  = ? kJ Question #3 is like question #1 in that we are given two equations and then asked to find out  H for a third equation. This one has a little problem. Look at the target equation and then look at the original equations. Half of the stuff in the target equation appears on the wrong side up in equation .

3. Examine the two equations below. 2Al(s) + 3/2 O 2 (g)  Al 2 O 3 (s)  H = kJ 2Fe(s) + 3/2 O 2 (g)  Fe 2 O 3 (s)  H = kJ Assuming that the above information is true, what is  H for the following equation? 2Al(s) + Fe 2 O 3 (s)  2Fe(s) + Al 2 O 3 (s)  H  = ? kJ What do we do?

 : Fe 2 O 3 (s)  2Fe(s) + 3/2 O 2 (g)  H = kJ 2Fe(s) + 3/2 O 2 (g)  Fe 2 O 3 (s)  H = kJ 3. Examine the two equations below. 2Al(s) + 3/2 O 2 (g)  Al 2 O 3 (s)  H = kJ Assuming that the above information is true, what is  H for the following equation? 2Al(s) + Fe 2 O 3 (s)  2Fe(s) + Al 2 O 3 (s)  H  = ? kJ Flip equation  Later you’ll add it to equation . Yeah. You can do that. HOWEVER, if you flip an equation You have to change the sign on the  H for that equation. You did just turn it into its opposite, so it’s only fair. Notice how I’m referring to the flipped equation as “  ”. The negative sign means it’s the opposite of equation .

 : Fe 2 O 3 (s)  2Fe(s) + 3/2 O 2 (g)  H = kJ 2Fe(s) + 3/2 O 2 (g)  Fe 2 O 3 (s)  H = kJ 3. Examine the two equations below. 2Al(s) + 3/2 O 2 (g)  Al 2 O 3 (s)  H = kJ Assuming that the above information is true, what is  H for the following equation? 2Al(s) + Fe 2 O 3 (s)  2Fe(s) + Al 2 O 3 (s)  H  = ? kJ Add the flipped verion of equation  to the normal version of equation .  : 2Al(s) + 3/2 O 2 (g) + Fe 2 O 3 (s)  2Fe(s) + 3/2 O 2 (g) + Al 2 O 3 (s) Cross out the oxygen gas term that appears on both sides of the arrow.

 : Fe 2 O 3 (s)  2Fe(s) + 3/2 O 2 (g)  H = kJ 2Fe(s) + 3/2 O 2 (g)  Fe 2 O 3 (s)  H = kJ 3. Examine the two equations below. 2Al(s) + 3/2 O 2 (g)  Al 2 O 3 (s)  H = kJ Assuming that the above information is true, what is  H for the following equation? 2Al(s) + Fe 2 O 3 (s)  2Fe(s) + Al 2 O 3 (s)  H  = ? kJ Time to clean up this town...  : 2Al(s) + 3/2 O 2 (g) + Fe 2 O 3 (s)  2Fe(s) + 3/2 O 2 (g) + Al 2 O 3 (s)  : 2Al(s) + Fe 2 O 3 (s)  2Fe(s) + Al 2 O 3 (s)  H  =  H  + (  H  ) = ( kJ) + ( kJ)= kJ

 : Fe 2 O 3 (s)  2Fe(s) + 3/2 O 2 (g)  H = kJ 2Fe(s) + 3/2 O 2 (g)  Fe 2 O 3 (s)  H = kJ 3. Examine the two equations below. 2Al(s) + 3/2 O 2 (g)  Al 2 O 3 (s)  H = kJ Assuming that the above information is true, what is  H for the following equation? 2Al(s) + Fe 2 O 3 (s)  2Fe(s) + Al 2 O 3 (s)  H  = ? kJ  : 2Al(s) + 3/2 O 2 (g) + Fe 2 O 3 (s)  2Fe(s) + 3/2 O 2 (g) + Al 2 O 3 (s)  : 2Al(s) + Fe 2 O 3 (s)  2Fe(s) + Al 2 O 3 (s)  H  =  H  + (  H  ) = ( kJ) + ( kJ)= kJ choice (C)

I haven’t finished making detailed explanations for all the questions yet! Here’s a strategy summary: #1: just add  &  together as is. Simple as it gets. #2: Flip  to get diamond on the right & add it to . #3: Flip  & add it to . #4: Add  +  + . Add  H’s & then multiply new  H by 7 #5: Multiply  by 2, Flip , then add modified  H’s (2  H  –  H  ). #6: flip , 3 x , 2 x , then add modified  H’s. (-  + 3  + 2  ) #7: ignore , add  & . Yeah.  was useless.

Click a button. Go to a place.™ Title page #1 lengthy explanation #2 #3 lengthy explanation #4 #5 #6 #7 #1 #3