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Calculating K from manipulations of Equation

Posted: Fri Nov 24, 2017 9:44 am
by Alex Kashou
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?

Re: Calculating K from manipulations of Equation

Posted: Fri Nov 24, 2017 12:04 pm
by Brigitte Phung 1F
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!