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.

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) + Cl2(g) SnCl2(s) H = -325.1 kJ

1. Consider the following equations:

SnCl2(s) + Cl2(g) SnCl4(l) H = -186.2 kJ

If the above is true, what is H for the following equation?

Sn(s) + 2Cl2(g) SnCl4(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) + Cl2(g) SnCl2(s) H = -325.1 kJ





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) + Cl2(g) SnCl2(s) H = -325.1 kJ





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) + Cl2(g) SnCl2(s) H = -325.1 kJ





Let’s do this.

Sn(s) + Cl2(g) SnCl2(s) H = -325.1 kJ





H III = ? kJDeclare your

unknown. This is your mission statement.

I

II

III

Sn(s) + Cl2(g) SnCl2(s) H = -325.1 kJ





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) + Cl2(g) SnCl2(s) H = -325.1 kJ





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) + Cl2(g) SnCl2(s) H = -325.1 kJ





H III = ? kJ

I + II: Sn(s) + Cl2(g) + SnCl2(s) + Cl2(g) SnCl2(s) + SnCl4(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) + Cl2(g) SnCl2(s) H = -325.1 kJ





H III = ? kJ


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) + Cl2(g) SnCl2(s) H = -325.1 kJ





H III = ? kJ


We can combine them.

Notice that Cl2(g) appears twice

on the left of the arrow.

Let’s clean up this equation. Time to re-write it!

Sn(s) + Cl2(g) SnCl2(s) H = -325.1 kJ





H III = ? kJ


The two separate Cl2(g)’s combine to form 2Cl2(g).

Also, the SnCl2(s)’s disappear.

I + II: Sn(s) + 2Cl2(g) SnCl4(l)

Sn(s) + Cl2(g) SnCl2(s) H = -325.1 kJ





H III = ? kJ


Look familiar?I + II: Sn(s) + 2Cl2(g) SnCl4(l)

Equation & combine to make equation III.

Sn(s) + Cl2(g) SnCl2(s) H = -325.1 kJ





H III = ? kJ



BIG IDEA: If equations I & comine to give us equation . . .

HHH

. . . You can add their H’s to get the H of equation

= (-325.1 kJ) + (-186.2 kJ)

H = -511.3 kJ = answer choice d.

Sn(s) + Cl2(g) SnCl2(s) H = -325.1 kJ





H III = ? kJ



HHH

= (-325.1 kJ) + (-186.2 kJ)

H = -511.3 kJ = answer choice d.

[phew]

3. Examine the two equations below.

2Al(s) + 3/2 O2(g) Al2O3(s) H = -1669.8 kJ

2Fe(s) + 3/2 O2(g) Fe2O3(s) H = -824.2 kJAssuming that the above information is true, what is H for the following equation?

2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)

H = ? kJQuestion #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 .


2Al(s) + 3/2 O2(g) Al2O3(s) H = -1669.8 kJ

2Fe(s) + 3/2 O2(g) Fe2O3(s) H = -824.2 kJAssuming that the above information is true, what is H for the following equation?


H = ? kJ

What do we do?

: Fe2O3(s) 2Fe(s) + 3/2 O2(g) H = + 824.2 kJ

2Fe(s) + 3/2 O2(g) Fe2O3(s) H = -824.2 kJ


2Al(s) + 3/2 O2(g) Al2O3(s) H = -1669.8 kJ

Assuming that the above information is true, what is H for the following equation?


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 .

: Fe2O3(s) 2Fe(s) + 3/2 O2(g) H = + 824.2 kJ

2Fe(s) + 3/2 O2(g) Fe2O3(s) H = -824.2 kJ


2Al(s) + 3/2 O2(g) Al2O3(s) H = -1669.8 kJ



H = ? kJ

Add the flipped verion of equation to the normal

version of equation .

:

2Al(s) + 3/2 O2(g) + Fe2O3(s) 2Fe(s) + 3/2 O2(g) + Al2O3(s)

Cross out the oxygen gas term that appears on both sides of the arrow.

: Fe2O3(s) 2Fe(s) + 3/2 O2(g) H = + 824.2 kJ

2Fe(s) + 3/2 O2(g) Fe2O3(s) H = -824.2 kJ


2Al(s) + 3/2 O2(g) Al2O3(s) H = -1669.8 kJ



H = ? kJ Time to clean up this town . . .

:


: 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)

H = H + (H)= (-1669.8 kJ) + (+824.2 kJ) = - 845.6 kJ

: Fe2O3(s) 2Fe(s) + 3/2 O2(g) H = + 824.2 kJ

2Fe(s) + 3/2 O2(g) Fe2O3(s) H = -824.2 kJ


2Al(s) + 3/2 O2(g) Al2O3(s) H = -1669.8 kJ



H = ? kJ

:


: 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)

H = H + (H)= (-1669.8 kJ) + (+824.2 kJ) = - 845.6 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 (2H – 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

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.

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