# Uncertainty of Method

1. Jan 8, 2005

Hello all

I was just wondering whether there is any systematic way in manipulating chemical formulas using Hess's Law to come up with the desired equation. Do you always have to guess and check Is it possible to use algebra and linear algebra?

Thanks

2. Jan 8, 2005

any ideas?

3. Jan 8, 2005

### Gokul43201

Staff Emeritus
There will be no systematic way if you are not told which equations to use (this is true of real life problems but not often of homework problems), to get the desired overall equation.

For nearly every homework problem I've seen, it's easy enough figuring it out by inspection (of course, you want to be smart about this, rather than randomly trying combinations), that using some algorithm might actually be slower.

That said, what you want is simply the values of $k_i$ such that
$$f(a,b,c,d,...,n) = \sum _i k_i f_i(a,b,c,d,...,n)$$
This simply results in a set of n simultaneous equations in the coefficients $k_i$, which may be solved by your chosen method (straight algebraic manipulation, using determinants, etc.)

Example (an easy one for simplicity, and I'm dispensing with details of the states, and such) :

We want $$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O~, ~~\Delta H_{comb}$$

given

$$C + 2H_2 \rightarrow CH_4~,~~\Delta H_1$$
$$C + O_2 \rightarrow CO_2~,~~\Delta H_2$$
$$2H_2 + O_2 \rightarrow 2H_2O~,~~\Delta H_3$$

Assume that

$$\Delta H_{comb} = k_1 \Delta H_1 + k_2 \Delta H_2 + k_3 \Delta H_3$$

Then, comparing terms we have :

$$from~~CH_4~:~1 = -k_1~~~~~(1)$$

$$from~~H_2~:~0 = 2k_1 + 2k_3 ~~~~~(2)$$

$$from~~C~:~0 = k_1 + k_2 ~~~~~(3)$$

$$from~~O_2~:~2 = k_2 + k_3 ~~~~~(4)$$

$$from~~H_2O~:~-2 = -2k_3 ~~~~~(5)$$

$$from~~CO_2~:~-1 = -k_2 ~~~~~(6)$$

Notice how I use +ve signs for terms on the LHS and -ve signs for terms on the RHS.

Anyway, we have 6 equations with 3 unknowns, of which, we suspect the other 3 will be redundant (but we shall verify this).

(1) tells us that $k_1 = -1$. Substituting this in (3) gives $k_2 = 1$. And (5) gives $k_3 = 1$. Clearly, these values also satisfy (2) (4) and (6), so they are consistent.

Hence, $$\Delta H_{comb} = - \Delta H_1 + \Delta H_2 + \Delta H_3$$

Of course, this would have be easier done by inspection, but this shows that an approach exists that will work well for the more complex problems, perhaps.

Last edited: Jan 8, 2005
4. Jan 8, 2005

Thanks a lot. But aren't you given the value of ∆H1 ∆H2 and ∆H3? I am not sure how you got the value of k1 k2 and k3.

Any help is appreciated

Thanks!

5. Jan 8, 2005

### Gokul43201

Staff Emeritus
Yes, you are given the values of ∆H1 ∆H2 and ∆H3, from which you want to find the value of ∆H(comb).

What don't you understand about the method used for determining the values of k1, k2 and k3 ?

Drat, I thought the explanation was understandable, but apparently not. Let me try and improve on it. But first, I need to understand the source of your confusion.

Do you agree that the objective is to find ∆H(comb) in terms of ∆H1, ∆H2 and ∆H3 ? Do you agree that the way to do this involves finding coefficients that multiply each of the given equations so that the sums or differences will result in the required equation ? Does it make sense that by multiplying by a -ve coefficient, you are simply reversing the direction of the reaction ?

Where exactly are you having a problem with my approach ?

6. Jan 8, 2005

like when you say:

from CH4: a = -k1
from H2: 0 = 2k1 + 2k3

etc.. how do we know this without knowing ∆H1 ∆H2 and ∆H3? Did we just look them up in the table?

Thanks a lot

7. Jan 8, 2005

### Gokul43201

Staff Emeritus
Typically, you know the values of ∆H1 ∆H2 and ∆H3. Either they will be given to you, in the problem, or you can look them up in a table.

As for the determination of the values of k, I was merely comparing coefficients of the different compounds between the parent reaction and the sub-reactions (which have been multiplied by k1, k2, k3 respectively).

In other words, compare the coefficients in : $$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$

with those in the sum of :

$$(k_1)C + 2(k_1)H_2 \rightarrow (k_1)CH_4$$,
$$(k_2)C + (k_2)O_2 \rightarrow (k_2)CO_2$$ and
$$2(k_3)H_2 + (k_3)O_2 \rightarrow 2(k_3)H_2O$$

The coeffecient of $CH_4$ in the main equation is 1. Its coefficient in the sum in -k1. Hence 1 = -k1.

Now since $H_2$ does not appear in the main equation, its coefficient there is 0. But it appears twice in the sum, with coefficients 2k1 and 2k3. So, 0 = 2k1 + 2k3.

Did that help ?

8. Jan 8, 2005