Calculating K from manipulations of Equation

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Alex Kashou
Posts: 38
Joined: Fri Sep 29, 2017 7:07 am

Calculating K from manipulations of Equation

Postby Alex Kashou » Fri Nov 24, 2017 9:44 am

Why do we square our K when we double our chemical equation? Similarly, why do we square root the K when we halve it as well?

Brigitte Phung 1F
Posts: 50
Joined: Thu Jul 27, 2017 3:00 am
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Re: Calculating K from manipulations of Equation

Postby Brigitte Phung 1F » Fri Nov 24, 2017 12:04 pm

We square our K value when we double the chemical equation because of how doubling the coefficients also doubles the exponents in the formula K = [product]^(product coefficient) / [reactant]^(reactant coefficient). For example, if you had 2A -> B, K = [B]^(1) / [A]^(2). When you double the equation, it becomes 4A - > 2B and then K = [B]^(2) / [A]^(4), which equals the original K squared. The same math happens when the equation is halved; the halved coefficients also result in halved exponents, which causes the new K to be the square root of the old K. Hope this helps!


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