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Chemistry A few questions, need help please

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TSCKC_1
post Jul 19 2009, 12:49 AM, updated 17y ago

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I have a few questions in my chemistry book that I do not understand enough to move on to the next section . Can someone help me?

Reactions rates are affected by concentration ,collision geometry , and the presence of a catalyst . Which of the following statement is false concerning these factors.

a Increasing the concentration of reacting particles increases the chance for collision
b Optimum collision geometry lowers the activation energy barrier
c A reaction occurs each time particles of the reactants collide.
d The slowest reaction incolved in a reaction mechanism determines the rate of overall reaction


the correct answer is c ,but why? I did alittle research online and found this statement "A chemical reaction occurs when substances (the reactants) collide with enough energy to rearrange to form different compounds (the products)."
isnt d correct" why then is c false?

Can someone also explain B and D?



Another question :

I needed to do this calculation :
I needed to balance the equation first (which i did already) and calculate ΔH
2(C2H6) + 7(O2) ---> 4(CO2) + 6(H2O) <<already balanced

C-H = 99kcal
O-O = 119kcal
C-C = 83.1kcal
C=O = 192kcal
O-H = 111kcal

ΔH = ??

I did the calculation this way
Products = C-H 2( 99*6) + C-C 2(83.1) + O-O 7(119) = 2187.2 kcal
Reactants = O=C 4(192*2) + H-O 6(111*2) = -2868

2187.2 + (-2868) = -680.8 kcal

The correct answer was -340.4 kcal
Why do i have to divide by 2?





TSCKC_1
post Jul 19 2009, 02:20 PM

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thank you so much , do you know how to do the other one?
TSCKC_1
post Jul 19 2009, 05:28 PM

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uhh , no , theres a graph in my book but i dont know how to draw the whole thing out..theres no O=O
TSCKC_1
post Jul 19 2009, 08:59 PM

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Here is the question

user posted image

QUOTE
  Pretty sure that you need to consider O=O though i maybe wrong because i'm pretty rusty with this.

Else, check the question, are they asking for Standard Enthalpy change of combustion or just enthalphy change of the reaction, if they are asking for standard enthalpy, means your jsut 1 step behind. Because standard enthalpy is defined as change per 1 mol of ethane. So just divide the energy change with the amount of mols of ethane which is 2.


I have to divide by 2 because the reaction requires 2 moles of C2H6 (ethane) right? how bout the 7 moles of Oxygen? does that affect the enthalpy in any way?

This post has been edited by CKC_1: Jul 19 2009, 09:04 PM
TSCKC_1
post Jul 19 2009, 09:27 PM

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okay got it now , thank you so much!

 

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