Chemistry Reaction Rates

  • #1
I came across this question on one of the practice papers I have and isn't sure about the answer, any help would be appreciated.

The following reaction occurs at constant temperature and constant volume in a closed system:

CaCO3(s) + 2H(ion)(aq) + 2Cl(ion)(aq) -> CO2(g) + H2O(l) + Ca(ion)(aq) + 2Cl(ion)(aq)

Changing which of the following would be useful in experimentally measuring the rate of this reaction?

a) The mass of the system

b) The pressure of the system

c) The concentration of water

d) the concentration of Cl ions

I think that the answer should be b), since c) and d) is obviously incorrect. Water is the product, so increasing its concentration doesn't help to determine the rxn rate. Cl ions cancel out at either end, and I don't know what the heck mass of the entire system means in terms of the reaction. Does it mean increasing the mass of every substance in the equation? Plz help.
 
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Answers and Replies

  • #2
Gokul43201
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Au contraire, the correct answers are (c) and (d) !
 
  • #3
Er, how can it? the chlorine ions exist on both sides, and they don't take part in the reaction itself. Water molecules exists only on the product side. There's also only 1 correct answer.
 
  • #4
loseyourname
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The correct answer is B. Changing the pressure will change the concentration of [itex]CO_{2} (g)[/itex]. Since the rate of the reaction can be expressed as
[tex]\frac{\Delta [CO_{2}]}{\Delta t}[/tex]
changing [itex][CO_{2}][/itex] will change the rate. C will change the equilibrium point of the reaction, but not the rate.
 
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  • #5
Gokul43201
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Oops ! I've bungled.

Yes, B will change the reaction rate. I did not notice a gaseous species. C and D are not correct for exactly the reasons you pointed out.

Sorry for the screw up !
 
  • #6
Yeah, i thought it was b) too. The main problem I'm having is understanding what increasing the mass of the "system" means. Can anyone explain? Does it mean increasing the mass of the entire reaction, such as all of the substances, etc.?
 
  • #7
Gokul43201
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Yes, that could be one interpretation. But since it is a wrong option, I wouldn't break my head over what the author/teacher meant by it.
 
  • #8
Huh? What do you mean by that?
 

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