Equilibrium Constant: H2+O2->H2O

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Discussion Overview

The discussion revolves around the equilibrium constant for the reaction of hydrogen and oxygen to form water, specifically addressing how the equilibrium constant is affected when the reaction equation is multiplied by a factor of two. The scope includes theoretical considerations of chemical equilibrium and mathematical reasoning related to constants.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant asks whether the equilibrium constant changes or remains constant when the reaction equation is multiplied by 2.
  • Another participant asserts that the square of a constant is still a constant, implying that the equilibrium constant may not change in a straightforward manner.
  • A subsequent reply seeks clarification on the previous statement regarding the constant.
  • Another participant argues that the equilibrium constant will change, but notes that the correct combination of the equation and constant will yield identical results in equilibrium calculations.
  • A further clarification is provided, suggesting that if the equilibrium constant for the original reaction is K, then for the modified equation, the new constant would be K², but this should not affect intensive or molar quantities.

Areas of Agreement / Disagreement

Participants express differing views on whether the equilibrium constant changes when the reaction equation is altered. Some participants suggest it remains constant in a certain sense, while others argue it does change, indicating a lack of consensus.

Contextual Notes

There are unresolved assumptions regarding the definitions of the equilibrium constant and how it relates to changes in the reaction equation. The discussion does not clarify the implications of these changes on equilibrium calculations.

Entanglement
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H2+1/2 O2 ---------> H20

If the equation is multiplied by 2
Will the equilibrium constant change or remain constant ?
 
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The square of a constant is still a constant.
 
Useful nucleus said:
The square of a constant is still a constant.
Sorry, I don't understand, can you clarify what you mean, thanks.
 
Equilibrium constant will change, but as it will describe different reaction equation, you will find when the correct combination of the equation and constant is used, results of equilibrium calculations are identical.
 
ElmorshedyDr said:
Sorry, I don't understand, can you clarify what you mean, thanks.
Suppose you calculate a constant for the above reaction and its value is K. If you multiply the equation by 2, then the new constant will be K2 .

But this should not matter for any intensive or molar quantity.
 

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