ive never truly understood how to draw hess cycles for reactions.

i have to do one for the decomposition of NaHCO3, and ive got this far -

1) NaHCO_{3} + HCl ---> NaCl + H_{2}O + CO_{2}

2) Na_{2}CO_{3} + 2HCl ---> 2NaCl + H_{2}O + CO_{2}

so if you multiply the NaHCO_{3} reaction by 2, and subtract the Na_{2}CO_{3} reaction, you end up with -

3) NaHCO_{3} ---> Na_{2}CO_{3} + H_{2}O + CO_{2}

so delta H for reaction 2 -(delta H for reaction 1 x 2) = delta H for reaction 3, no?

where do i go from here to construct a hess cycle? is what ive done even correct?

(sorry if its badly expressed or confusing)

i have to do one for the decomposition of NaHCO3, and ive got this far -

1) NaHCO

2) Na

so if you multiply the NaHCO

3) NaHCO

so delta H for reaction 2 -(delta H for reaction 1 x 2) = delta H for reaction 3, no?

where do i go from here to construct a hess cycle? is what ive done even correct?

(sorry if its badly expressed or confusing)

Slazenger

ive never truly understood how to draw hess cycles for reactions.

i have to do one for the decomposition of NaHCO3, and ive got this far -

1) 2NaHCO_{3} + 2HCl ---> 2NaCl + 2H_{2}O + 2CO_{2}

2) Na_{2}CO_{3} + 2HCl ---> 2NaCl + H_{2}O + CO_{2}

so if you multiply the NaHCO_{3} reaction by 2, and subtract the Na_{2}CO_{3} reaction, you end up with -

3) 2NaHCO_{3} ---> Na_{2}CO_{3} + H_{2}O + CO_{2}

so delta H for reaction 2 -(delta H for reaction 1 x 2) = delta H for reaction 3, no?

where do i go from here to construct a hess cycle? is what ive done even correct?

(sorry if its badly expressed or confusing)

i have to do one for the decomposition of NaHCO3, and ive got this far -

1) 2NaHCO

2) Na

so if you multiply the NaHCO

3) 2NaHCO

so delta H for reaction 2 -(delta H for reaction 1 x 2) = delta H for reaction 3, no?

where do i go from here to construct a hess cycle? is what ive done even correct?

(sorry if its badly expressed or confusing)

equation 1 (times 2): 2NaHCO

equation 2: Na

you want the NaHCO3 on the left hand side so your sum is:

(2 x equation 1) - equation 2 = required equation

rearrange

required equation + equation 2 = (2 x equation 1)

to make a Hess cycle from this:

you can see that if you add required equation to equation 2 you get (2 x equation 1)

.

Thus the arrows must go from required equation to equation 2 to (2 x equation 1)

(the indirect route)

and directly from required equation to (2 x equation 1)

(the second route)

checksum:

required equation + equation 2

2NaHCO

+

Na

--------------------------------------------------------------------

2NaHCO

which is 2 x equation 1

I got so confused with the moles etc with this Q that I gave up

But the Hess's cycle has to have the same products-just difference in moles: the Bicarbonate has to be x 2 when you work out the number of moles. So instead of dividing the mass you have by only 84 you divide it by 168.

then what you do is:

You use the equation for the Hess's cyle:

* D stands for the decomposition

DHD + DH1 --> DH2

Adding in your values:

DHD -29.3 = 43.6

(for example!)

Then rearrange to get

DHD=43.6 + 29.3

= 72.9 kJmol^-1

If that confuses you feel free to ask.

But the Hess's cycle has to have the same products-just difference in moles: the Bicarbonate has to be x 2 when you work out the number of moles. So instead of dividing the mass you have by only 84 you divide it by 168.

then what you do is:

You use the equation for the Hess's cyle:

* D stands for the decomposition

DHD + DH1 --> DH2

Adding in your values:

DHD -29.3 = 43.6

(for example!)

Then rearrange to get

DHD=43.6 + 29.3

= 72.9 kJmol^-1

If that confuses you feel free to ask.

charco

equation 1 (times 2): 2NaHCO_{3} + 2HCl ---> 2NaCl + 2H_{2}O + 2CO_{2}

equation 2: Na_{2}CO_{3} + 2HCl ---> 2NaCl + H_{2}O + CO_{2}

you want the NaHCO3 on the left hand side so your sum is:

(2 x equation 1) - equation 2 = required equation

rearrange

required equation + equation 2 = (2 x equation 1)

to make a Hess cycle from this:

you can see that if you add required equation to equation 2 you get (2 x equation 1)

.

Thus the arrows must go from required equation to equation 2 to (2 x equation 1)

(the indirect route)

and directly from required equation to (2 x equation 1)

(the second route)

checksum:

required equation + equation 2

2NaHCO_{3} ---> Na_{2}CO_{3} + H_{2}O + CO_{2}

+

Na_{2}CO_{3} + 2HCl ---> 2NaCl + H_{2}O + CO_{2}

--------------------------------------------------------------------

2NaHCO_{3} + 2HCl ---> 2NaCl + 2H_{2}O + 2CO_{2}

which is 2 x equation 1

equation 2: Na

you want the NaHCO3 on the left hand side so your sum is:

(2 x equation 1) - equation 2 = required equation

rearrange

required equation + equation 2 = (2 x equation 1)

to make a Hess cycle from this:

you can see that if you add required equation to equation 2 you get (2 x equation 1)

.

Thus the arrows must go from required equation to equation 2 to (2 x equation 1)

(the indirect route)

and directly from required equation to (2 x equation 1)

(the second route)

checksum:

required equation + equation 2

2NaHCO

+

Na

--------------------------------------------------------------------

2NaHCO

which is 2 x equation 1

thanks for such a full explanation and the help appreciated

Malsi101

I got so confused with the moles etc with this Q that I gave up

But the Hess's cycle has to have the same products-just difference in moles: the Bicarbonate has to be x 2 when you work out the number of moles. So instead of dividing the mass you have by only 84 you divide it by 168.

then what you do is:

You use the equation for the Hess's cyle:

* D stands for the decomposition

DHD + DH1 --> DH2

Adding in your values:

DHD -29.3 = 43.6

(for example!)

Then rearrange to get

DHD=43.6 + 29.3

= 72.9 kJmol^-1

If that confuses you feel free to ask.

But the Hess's cycle has to have the same products-just difference in moles: the Bicarbonate has to be x 2 when you work out the number of moles. So instead of dividing the mass you have by only 84 you divide it by 168.

then what you do is:

You use the equation for the Hess's cyle:

* D stands for the decomposition

DHD + DH1 --> DH2

Adding in your values:

DHD -29.3 = 43.6

(for example!)

Then rearrange to get

DHD=43.6 + 29.3

= 72.9 kJmol^-1

If that confuses you feel free to ask.

the maths part i understand, it was just the hess cycle that i had problems with.

I have this really simple but daft model that I use with me students but they seem to get the drift quite quickly. I draw an outline of Britain, put a point at Bolton (the social, education and evolutionary centre of excellence on the UK) and London (an overpriced hell). That distance is fixed, but then we go differnet routes, and no matter how far we travel (eg via inverness, or norwich etc) the distance from our start point is always the same.

Then I get them to put the reaction they are trying to find on the map - and we use the data they have to make the connections between them. Then we find a route from the start point to the destination - and if we go against the arrow we change the sign and, with the arrow we keep the sign!)

Only takes a few goes ut you'll get it quickly enough

Rob

Then I get them to put the reaction they are trying to find on the map - and we use the data they have to make the connections between them. Then we find a route from the start point to the destination - and if we go against the arrow we change the sign and, with the arrow we keep the sign!)

Only takes a few goes ut you'll get it quickly enough

Rob

chemicalguy

I draw an outline of Britain, put a point at Bolton (the social, education and evolutionary centre of excellence on the UK) and London (an overpriced hell). That distance is fixed, but then we go differnet routes, and no matter how far we travel (eg via inverness, or norwich etc) the distance from our start point is always the same.

This is a better model for explaining distance vs. displacement? The amount of petrol you would use driving to Inverness and then back down wouldn't be the same as a direct route and so the energy would be different

(Though I understand it's just a learning aid)

Slazenger

just out of interest, should this be an exothermic or an endothermic reaction?

ive tried to find a texbook value as to what it should be, but ive failed miserably!

ive tried to find a texbook value as to what it should be, but ive failed miserably!

The reaction you need is a decomposition. This requires breaking something apart and it is likely that it needs energy input to do so. So it is endothermic.

Original post by Slazenger

good! thanks

i got an answer of: +82.1 kJmol^{-1}

i got an answer of: +82.1 kJmol

The answer 82.1 kJmol^-1, is that the data/ text book value or just the one you got from experiments? I’ve scoured the internet for the data book delta H for the overall decomposition of NaHCO3 -> Na2CO3 + H2O + CO3

Original post by Tigress6281

The answer 82.1 kJmol^-1, is that the data/ text book value or just the one you got from experiments? I’ve scoured the internet for the data book delta H for the overall decomposition of NaHCO3 -> Na2CO3 + H2O + CO3

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