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Topic: Enthalpy:2Mg(s) + O2(g) --> 2MgO(s) ?H = -1204  (Read 47939 times)

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Offline JZ_1

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Enthalpy:2Mg(s) + O2(g) --> 2MgO(s) ?H = -1204
« on: October 23, 2005, 01:41:18 AM »
It says consider the reaction:

2Mg(s) + O2(g) --> 2MgO(s)   ?H =  -1204

1) Calculate the amount of heat transfered when 2.4 grams of Mg(s) reacts at constant pressure

2) How many grams of MgO are produced during an enthalpy of 96.0kJ?

3) How many kilojoules of heat are absorbed when 7.50 g of MgO(s) are decomposed into Mg and O2(g) at constant pressure

====================================
for number 1
  by converting 2.4g Mg to moles and multiplying by ?H, and dividing by 2....I have -59kJ heat transfered

for number 2 and 3
  HOW DO I START???
  is there even enough information?

Somehow the book have 6.43 g MgO produced and +112kJ heat absorbed
  but how can you get 6.43 g MgO produced from the information in number 2?

you may need to define empathy....because all i know is that ?H is:  ? kJ per 1 mole of substance
« Last Edit: October 23, 2005, 07:01:57 PM by Mitch »
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Offline sdekivit

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Re:Enthalpy ...NEED *delete me*!
« Reply #1 on: October 23, 2005, 05:53:27 AM »
for 2 you need to use the stoichiometry of the reaction, thus you know that 1 mol Mg will produce 1 mol MgO

yopu know that 2 mol Mg that reacted releases 1204 kJ(or J --> no unit given) thus how many mol Mg releases 96 kJ ?

for 3 you need to know that the reverse reaction needs the enthalpychange that is released in the given reaction. Thus:

2MgO --> 2 Mg + O2 --> delta H = +1204 (what ?)

this is also per 2 mol MgO.
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Offline JZ_1

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Re:Enthalpy ...NEED *delete me*!
« Reply #2 on: October 23, 2005, 12:30:39 PM »
ok...
For number 2,  by dividing 96.0 kJ by -1204 kJ/mol and then multiplying it by 40.3 grams MgO....i got -3.21

for number 3,  by changing 7.50 g MgO to moles and multiplying it by 1204 kJ/mol i have 224KJ....

I want to ask why my answer of 3.21 is half the actual answer....also why 224kJ is 2x the actual answer..

Am i suppose to multiply the first one by 2?....why would i do that?...i thought it is a 1:1 ratio...

thank you in advance
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Offline sdekivit

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Re:Enthalpy ...NEED *delete me*!
« Reply #3 on: October 23, 2005, 12:39:10 PM »
alright here it goes:

2--> 2 mol MgO produced releases 1204 kJ thus 96 kJ produces (96 * 2)/1204 = 0,159468 mol MgO

--> thus 6,43 g MgO

3--> 7,5 g MgO = 0,186058 mol MgO.

2 mol MgO absorbes 1204 kJ thus (2 * 0,186058)/1204 = 112 kJ heat is absorbed
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Offline JZ_1

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Re:Enthalpy ...NEED *delete me*!
« Reply #4 on: October 23, 2005, 12:54:00 PM »
Quote from: sdekivit on October 23, 2005, 12:39:10 PM
3--> 7,5 g MgO = 0,186058 mol MgO.

2 mol MgO absorbes 1204 kJ thus (2 * 0,186058)/1204 = 112 kJ heat is absorbed

that doesn' t give you 112 KJ heat (shouldn't you multiply by 1204?)..but i think i know the concept...

Oh!..one more thing...why isn't number 6.43 a negative number?....aren't we suppose to divide it by -1204?
« Last Edit: October 23, 2005, 01:04:01 PM by JZ_1 »
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Offline constant thinker

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Re:Enthalpy ...NEED *delete me*!
« Reply #5 on: October 23, 2005, 07:00:53 PM »
I think you guys need an Enthalpy board.
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Offline xiankai

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Re:Enthalpy:2Mg(s) + O2(g) --> 2MgO(s) ?H = -1204
« Reply #6 on: October 24, 2005, 08:34:31 AM »
mass cannot be negative.

we dont divide it by -1204 because we're finding the raw quantity of energy change.

remember, energy change and enthalpy change are different.

energy change is finding out how much energy was changed, regardless of whether it went down or up. u dont say that the energy decreased by -1204 kJ.
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Offline sdekivit

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Re:Enthalpy ...NEED *delete me*!
« Reply #7 on: October 24, 2005, 11:13:24 AM »
Quote from: JZ_1 on October 23, 2005, 12:54:00 PM
that doesn' t give you 112 KJ heat (shouldn't you multiply by 1204?)..but i think i know the concept...

i said it absorbes 112 kJ heat. Remember:

Enthalpy of formation = - Enthalpy of decomposition

The sign is positive as I look from within the system. This absorbes energy so there is an increase of energy in the system --> positive sign.
« Last Edit: October 24, 2005, 11:14:35 AM by sdekivit »
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