How to calculate delta H ... Please Help!?

Question

Given the following reactions and delta H values:

B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) delta H = +2035 kJ

2H2O(l) ----> 2H2O(g) delta H = +88 kJ

H2(g) +0.5O2(g) ----> H2O(l) delta H = -286 kJ

2B(s) +3H2(g) ---->B2H6(g) delta H = +36 kJ

Calculate delta H for

2B(s) + 1.5O2(g) ----> B2O3(s) delta H = ?

Please Explain How To Solve The Problem. I tried to rearrange the formulas over and over again but I am not getting the correct answer (which is -1273 kJ).

Answer 1

for this equation( B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g))to happen,

the following equations have to happen:

1- 2B(s) + 1.5O2(g) ----> B2O3(s) but in the reverse direction

2-  3H2O(l) ----> 3H2O(g) but in the reverse direction

3- 3H2(g) +1.5O2(g) ----> 3H2O(l) but in the reverse direction

4-2B(s) +3H2(g) ---->B2H6(g)

using the following rules:

a.when a reaction happens in the reverse dirrection as in equations 1-3  above, the deltaH is multiplied by minus one, and if an equation is multiplied by an integar number, then it's deltaH is also multiplied by that.

b.the sum of the deltaH of the 4 equations above add up to 2035.

we will name the first equation's deltaH,  when the reaction is done in this direction:2B(s) + 1.5O2(g) <---- B2O3(s) , X . so the answer to your question is -X because the reaction you asked is the above done in the opposite direction.

the second equations deltaH is -88 multiplied by 1.5(=132) . according to H2(g) +0.5O2(g) ----> H2O(l) being done in the other direction and multipliying its deltaH by 1.5 to have 3 H2O molecules.

the third equations deltaH is 3multiplied be 286(=858) according to H2(g) +0.5O2(g) ----> H2O(l) being done reverse and multiplied by 3.

the fourth equations deltaH is 36.

so adding the delta H of the equations up and putting it equal to 2035 we have:

X-132+858 + 36 = 2035

X=1273

so the answer is -1273.

Answer 2

I received your e-mail asking me to do a walk through this hess's law problem. yes it is more envolved than the typical problem, & having a metod to approach such a problem helps.

you have right answer with the top person, but here  is the "sole provider" method,...

(A) B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ……delta H = +2035 kJ

(B) 2H2O(l) ----> 2H2O(g) ……….   ……..   ……..   ..delta H = +88 kJ

(C) H2(g) +0.5O2(g) ----> H2O(l)  ÃƒÂ¢Ã‚€Â¦Ã¢Â€Â¦.       ……..     delta H = -286 kJ

(D) 2B(s) +3H2(g) ---->B2H6(g)  ÃƒÂ¢Ã‚€Â¦Ã¢Â€Â¦..       ……..      delta H = +36 kJ

Calculate delta H for

2B(s) + 1.5O2(g) ----> B2O3(s) delta H = ?

You find equations that are “sole providers”

Use equation D is the only provider of the desired starting material Boron

“1D” :  2B(s) +3H2(g) ---->B2H6(g

Use the opposite “-A” BECAUSE IT IS THE ONLY PROVIDER OF B2O3, but we need it as a product & it provides it as a starting material

“-1A” :  B2H6(g) + 3O2(g) - B2O3(s) + 3H2O(g)  this is “A” reversed

Your combined set of 1D & -1A ,…so far is:

B2H6(g) + 3O2(g) & 2B(s) +3H2(g)  - B2O3(s) + 3H2O(g & B2H6(g)

Which simplifies to

3O2(g) & 2B(s) +3H2(g)  - B2O3(s) + 3H2O(g)

Now we clean up, by finding equations that are the sole providers of molecules that we wish eliminated, I.e. to get rid of  3H2O(g) products , I need 3H2O(g) as starting materials , … found by using “-1.5 B”  ( the reverse of 1.5 B’s)

1D & -1A :   3O2(g) & 2B(s) +3H2(g)  - B2O3(s) + 3H2O(g)

-1.5 B : 3 H2O(g) ----> 3H2O(l)

Which leaves us with:

1D & -1A  -1.5 B :   3O2(g) & 2B(s) +3H2(g)  - B2O3(s) + 3H2O(l)

Now to get rid of 3 H2O(l) products ,  we need 3 H2O(l) as starting materials found by using “-3C”

1D & -1A  -1.5 B :   3O2(g) & 2B(s) +3H2(g)  - B2O3(s) + 3H2O(l)

-3C  :  3 H2O(l)  ÃƒÂ¯Ã‚ƒÂ   3 H2(g) + 1.5O2(g)

You may notice that now everything now cancels out except for what you wanted:

2B(s) + 1.5O2(g) ----> B2O3(s)

& Hess’s Law states that if 1D & -1A  -1.5 B -3C got you what you wanted, then dHD – dHA -1.5dHB -3dHC will give you the energy that you want:

dHD – dHA -1.5dHB -3dHC =

+36 kJ – (+2035 kJ) – (1.5)( 88 Kj) – (3)( -286 kJ)

-1999 -132 + 858

dH = -1273kJ

p.s. (text book for this reaction  is -1272.77)

Answer 3

I think you know the basic idea behind the solution to this problem. You are given several reactions along with their delta H values and your goal is to manipulate the reactions and sum their delta H values to find the delta H of the given reaction. Sometimes manipulating the reactions can be tricky but it's really easy.

You have 4 reactions and I'm just going to tell you what you need to do to each reaction and them I'll try to explain why.

1. Reverse the first reaction and change the sign for delta H from +2035 kJ to -2035 kJ.

2. Reverse the reaction, multiply the entire reaction by 3/2, and change the delta H value from +88 kJ to -(3/2)*(88) kJ = -132 kJ.

3. Reverse the reaction, multiply the entire reaction by 3, and change the delta H value from -286 kJ to -3(-286) kJ = 858 kJ

4. This reaction remains unchanged.

If you add those numbers you will confirm -1273 kJ as the delta H for the overall reaction.

This is how I approach problems like this.

First I look at the form of the overall reaction. For this particular overall reaction there is 2B(s) and 1.5O2 (g) in the reactants, so automatically I know that I don't have to reverse reaction 4.

When I look at the product of the overall reaction I see that it has B2O3 (s) and since reaction 1 is the only reaction with this species I know automatically that I have to reverse the first reaction.

Upon further inspection I realize that there is no water or H2 (g) in the overall reaction so I will have the manipulate reactions 2 and 3 to remove both of those species.

You can only cancel out like species that are in the same state (there is both liquid water and gaseous in these reactions) and on opposite sides of two or more different reactions. To remove all of the gaseous water I have to reverse the second reaction so that the gaseous water in the products of reaction 1 cancels out the gaseous water in the reactants of reaction 2.

The only problem with this is that there are 3 H2O (g) in reaction 1 and only 2 in reaction 2. Since there is only 1 B2SO3 (s) in the overall reaction I know that I can't multiply reaction 1 by a factor, so the easiest way to remove the H2O (g) is the multiply reaction 2 by a factor of 3/2 because (3/2)*(2) = 3, which gives me exactly the same amount of water in both reactions 1 and 2.

As a result of reversing reaction 2, the H2O (l) in now is the products and since I want the H2O (l) in reaction 2 to cancel out the H2O (l) in reaction 3, I have to reverse reaction 3. There are 3 H2O(l) in reaction 2 and only 1 H2O (l) in reaction 3, so just multiply reaction 3 by 3 to get the same amount of water on both sides and cancel them out.

Finally just sum the overall reaction and compute the new delta H.