[Salerk]

To start, sorry if this is in the incorrect forum, i was just so unsure as to what one it may fit into, i decided general will work and hopefully a kind mod will move it to the correct area if necessary.

Due to a very bad flu i was out of class for 2 weeks, coming back Ive been asked how to do the following question and have no clue what's so ever, so rather than an answer I was wondering if anyone can simply explain how i am to solve this question.  Once I know the how I'm more than happy to do it myself, and unfortunately i wont be back in class for another week and a bit so unable to ask my tutor. so until then...

Consider the following reaction involving glucose.

C₆H₁₂O₆ +O₂  --> CO₂ + H₂O  Enthalpy (AH) = -2816

Write the balanced equation

is this endothermic or exothermic reaction

Calculate the energy produced or absorbed when 3.7mol of glucose burn with excess O2

Calculate the energy produced or absorbed when 12g of glucose burn with excess O2

What i know is that the balances equation will be.

C₆H₁₂O₆ + 6O₂ --> 6CO₂ + 6H₂O

no problems there (I hope).

First part I don't understand is if its an endothermic or exothermic.

Endo (ENter) Exo (EXit) OK. so its a negative, meaning i believe its releasing heat (would this be correct) making it and Endothermic reaction.

Question: because the equation has been balanced the numbers have changed, will this have altered the Enthalpy?  If so how do you calculate what its been changed to, if not why does it not change?

Assuming that it does not change (if it does I would just substitute in the new numbers)

You start with 1 mol of glucose, this is increased to 3.7 would you do the following?

1mol = -2816 thus 3.7 * 2816 = -10419.2KJ

if this process is right then the next one I'm fine with, but I just don't know if this is how the questions meant to be answered, any help would be REALLY great, thank you very much

Sal

p.s. Sorry for poor spelling, the built in spell check helped a lot tho!

[Borek]

Endo (ENter) Exo (EXit) OK. so its a negative, meaning i believe its releasing heat (would this be correct) making it and Endothermic reaction.

Quite the opposite. Heat exits the system, thus the reaction is exothermic.

Question: because the equation has been balanced the numbers have changed, will this have altered the Enthalpy?  If so how do you calculate what its been changed to, if not why does it not change?

Assume it is enthalpy per 1 mole of glucose burned. Sometimes it is called 'enthalpy per mole of reaction' (whatever it means )

You start with 1 mol of glucose, this is increased to 3.7 would you do the following?

1mol = -2816 thus 3.7 * 2816 = -10419.2KJ

OK

[Salerk]

Sorry, I was just wondering if you would explain what you mean by

Assume it is enthalpy per 1 mole of glucose burned. Sometimes it is called enthalpy per mole of reaction (whatever it means )

You start with 1 mol of glucose, this is increased to 3.7 would you do the following?

1mol = -2816 thus 3.7 * 2816 = -10419.2KJ

OK

and does OK mean that was correct? On lots of pain killers, so finding everything is a little of a blur right now *smile*

[Borek]

You are correct - OK means you were correct

Enthalpy you were given was not for the unbalanced equation, but for the complete combustion of glucose. Reaction was given in the form of skeletal equation, jut to signal what it means by complete combustion. Correct reaction equation is the one after balancing, and this balanced reaction gives 2816kJ per mole of glucose.

[Salerk]

To start I would like to say thank you very much for the reply would very helpfull! sorry ive been a while in my reply giving thanks.

I checked the rules and unless i missed it i believe its OK to post a separate question but an extention of the first in this thread, so hopfully OK to just be added into the same post, don't like posting to many times especially since the questions are so closely related.

OK here is the question

The oxidation of nitrogen in the hot exhaust of jet engines and automobiles occurs by the reaction
   N₂(g) + O₂(g)® NO(g)       DHo = +180.6 kJ
Write the balanced equation
Is this an endothermic or an exothermic reaction? Justify your answer
Calculate the energy produced (or absorbed) to form 3 mol of NO.
Calculate the energy produced (or absorbed) to form 2.7 g of NO.

ok to start, when reading it I thought at first was identical to first, but only with a change of formulae this was altered when I saw you need to calculate out how to from rather than reacted (burnt) . I theorised that since its the reverse you would do the reverse.

so...

Balanced its

N₂ + O₂ = 2NO


Is this an endothermic or an exothermic reaction? Justify your answer

The reaction is endothermic because heat is being taken into the reaction.

Calculate the energy produced (or absorbed) to form 3 mol of NO.

to form 2 mol of NO (2NO) you need +180.6KJ to form 3 you would need half that again, so

180.6 / 2 = 90.3
= 180.6 + 90.3 =270.9KJ

Calculate the energy produced (or absorbed) to form 2.7 g of NO.

NO = (N = 14.01) + (O = 16.0) = 30.1 MM

n = m / mm
n = 2.7/30.1
n = 0.090

180.6 * 0.090 = 16.2KJ of energy needed to form.

So, hopfully did OK 

[Borek]

NO = (N = 14.01) + (O = 16.0) = 30.1 MM

Close. Check your math

180.6 * 0.090 = 16.2KJ of energy needed to form.

No. Hint: 2NO.

[Salerk]

NO = (N = 14.01) + (O = 16.0) = 30.1 MM

Close. Check your math

180.6 * 0.090 = 16.2KJ of energy needed to form.

No. Hint: 2NO.

Ahh oops. so 30.0 => 30.01

subbed in = 0.090 (rounded)

as for the next bit if its to 2mol then I need to take it to 1 mol? and sub in the number?

so would it be 90.3 * 0.090 = 8.127 (8.13 KJ)

hopfully Ive done that right now.

[Salerk]

In the manufacture of nitric acid by the oxidation of ammonia, the first product is nitric oxide (NO), which is then oxidised to nitrogen dioxide (NO2). Using the standard reaction enthalpies
   N2(g) + O2(g) ? 2NO (g)    DHo = +180.6 kJ
   N2(g) + 2O2(g)? 2NO2 (g)    DHo = +66.4 kJ


Calculate the DHo for the following reaction
2NO (g) + O2(g) ? 2NO2 (g)

As far as I know the amounts of Enthalpy stay the same no matter what is going on with it, so

If the total of the components that make 2NO are +180.6
And the total that makes 2NO2 are +66.4

if I subtract the +180.6 and the +66.4 It should show me the enthalpy for the equation that is unknown on. Would this be correct?

= 114.2 Enthalpy

I will be honest in admitting this is almost a complete guess, because i'm really unsure on how about to go on with this question.

[charco]

reverse equation 1 and add it to equation 2 and it gives you the desired equation.

Do the same for the energies and you get -180.6 + 55.4 = -174.2 kJmol-1