Major problems with chemistry

if your teacher is not really in school where is he really?

Combustion of magnesium carbonate = + 100 kj/mol -1

errr - magnesium carbonate certainly does NOT combust

What is the standard entalpy of combustion of carbon:

2C (s) + O2 (g) = 2 CO (g) = -222 kj/mol -1
2CO (g) + O2 (g) = 2CO 2 (gO = -566 kj/mol -1

I have just added the two values together. Is that right?

no, if you add them together you get:
2C + 2O2 --> 2CO2
which is twice the value that you need

Calculate the molar enthalpy of hydrogenation of but-2-ene, given the following standard enthalpies of combustion:

C4H8 (g) = -2718
H2 (g) = -286
C4H10 (g) = -2877

What does the 'molar enthalpy' mean in this case.?

it means the enthalpy change per mole - nothing else, no complications. It means that you have to calculate it for only 1 mole of reactant

stage 1 : write out the equation that you need to calculate

CH3CH=CHCH3 + H2 --> CH3CH2CH2CH3

stage 2: construct this equation by manipulating the equations that you ARE given values for.

C4H8 + 6O2 --> 4CO2 + 4H2O delta H = -2718
C4H10 + 6.5O2 --> 4CO2 + 5H2O delta H = -2877
------------------------------------------ subtract and rearrange
C4H8 + H2O --> 0.5O2 + C4H10 ..............delta H = 2877 - 2718 = +159
H2 + 0.5O2 --> H2O ..........delta H = -286
------------------------------------------ add
C4H8 + H2 --> C4H10 ......... deltaH = +159 - 286 = -127 kJ mol-1

better check my sums ...

Reply 2

Thank you so much. He is away having an operation to his neck at the moment....so we are getting cover teachers, who just give worksheets and stuff. The topic has just not "clicked" in my head yet, if you know what I mean. I still haven't really figured it out. Thanks though! YOU GET REP! :biggrin:

Reply 3

Oh and sorry. It said the value for the decomposition of MgCO3 was + 100...how would I do the calculation then? Is what I did right? Thank you very much.

Amrou

Reply 4

Well the equation for decomposition of MgCO3 is:

MgCO3 --> MgO + CO2

.... which is exactly what you have been given AND asked to find - so there is something horribly wrong there!

Reply 5

Ooops my fault .... again. The question wants me to calculate the standard enthalpy of formation of magnesium carbonate....sorry.

Thank you very much

Amrou

Reply 6

standard enthalpy of formation is the energy change when 1 mole of a substance is formed from its elements in their states under standard conditions

Mg(s) + C(s) + 1.5O2(g) --> MgCO3 delta Hf

you have to construct this equation from the others that you have been given.

MgCO3 (s) = MgO (s) + CO2 (g) ........... delta H = +100 kJ

reverse it

MgO (s) + CO2 (g) --> MgCO3 (s) ........... delta H = -100 kJ
Mg + 0.5O2 --> MgO ........................... delta H = -602kJ
------------------------------------------ add together
Mg + CO2 + 0.5O2 ----> MgCO3 ............. delta H = -702 kJ
C + O2 ---> CO2 ..............................delta H = -394kJ
------------------------------------------- add together and cancel on both sides
Mg + C + 1.5O2 ---> MgCO3 ........... delta H = -1096 kJ (exothermic)

we agree numerically but disagree on the sign

Reply 7

Thank you very much. Where I struggle most is knowing when to subtract the products from the reactants, or the other way around.

One question says:

Ethyne burns in oxygen according to the equation:

2C2H2 + 5O2 = 4CO2 + 2H20 (l)

I have been given values for the standard enthalpy of formations.

I have just subtracted products from reactants? Was this right? Or is there another way of doing this?

Thanks alot. Your help is greatly appreciated.

Amrou

Reply 8

amrou

Thank you very much. Where I struggle most is knowing when to subtract the products from the reactants, or the other way around.
One question says:
Ethyne burns in oxygen according to the equation:
2C2H2 + 5O2 = 4CO2 + 2H20 (l)

I have been given values for the standard enthalpy of formations.
I have just subtracted products from reactants? Was this right? Or is there another way of doing this?

Like I said in a previous post, you just construct the equation that you want step by step. If you want to remove something from the left hand side you either add an equation that has the same quantity of that component on the right hand side.....
.... or you subtract an equation that has the same quantity of that component on the left hand side.
It may be that you have to multiply an equation to acheive the same quantity of the component but that's alright too.
-------------------------------------------------------------------------------------------------
However, if you have been given all of the enthalpies of formation there is a shorter way to do it...

To understand the method you must consider EXACTLY wht the enthalpy of formation means. It is the energy required to form 1 mole of a compound from its elements in their standard states.

Now if you do the reverse you get the elements back from the compound. i.e. the reverse enthalpy of formation (call it the enthalpy of reverse formation or whatever you like).

So if you are faced with an equation first of all take the reactants and reverse their enthalpies of formation (remembering that by definition the enthalpy of formation for an element is zero) - and then add the enthalpies of formatino of the products ('cos they have to be formed, right?)

Now, mathematically this is:
reaction enthalpy = (sum of the negative enthalpies of formation of the reactants) + (sum of the enthalpies of formation of the products)

If you rearrange the equation you get the form that is usually stated in textbooks:
Reaction enthalpy = (sum of the Enthalpy of formation of the products) - (sum of the enthalpy of formation of the reactants)

so in your example 2C2H2 + 5O2 = 4CO2 + 2H20 your sum is (remember oxygen is zero):

[4 x deltaHf(CO2)] + [2 x deltaHf(H2O)] - [2 x deltaHf(C2H2)] = enthalpy of combustion

Reply 9

paul_the_enigma

I have just been given the same sheet as above!

charco

If you rearrange the equation you get the form that is usually stated in textbooks:
Reaction enthalpy = (sum of the Enthalpy of formation of the products) - (sum of the enthalpy of formation of the reactants)

so in your example 2C2H2 + 5O2 = 4CO2 + 2H20 your sum is (remember oxygen is zero):

[4 x deltaHf(CO2)] + [2 x deltaHf(H2O)] - [2 x deltaHf(C2H2)] = enthalpy of combustion

.... Does the enthalpy of combustion not equal delta H Reactants - delta H products? .... like THIS? ... (the opposite way to above) ... as ethyne is a reactant? or isn't it??

[2 x deltaHf(C2H2)] - [4 x deltaHf(CO2)] + [2 x deltaHf(H2O)]

and so equal: 228 - (4x -394) + (2x -286) = 2376 kJ?

I may be very wrong! Some feedback would be greatly appreciated!

Ta! paul.

Reply 10