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### Changing Kc

Posted: **Sat Dec 09, 2017 3:11 pm**

by **juchung7**

In the practice final, question number 32 asks if Kc is x at 400 K, what is Kc if the stoichiometric coefficients are halved? The answer says the root of x or x^1/2, and I know this makes sense mathematically, but doesn't the value of Kc for a reaction only change based on temperature?

### Re: Changing Kc

Posted: **Sat Dec 09, 2017 3:27 pm**

by **Kyra Dingle 1B**

The value of Kc can change when stoichiometric coefficients are halved because when calculating Kc itself it requires the stoichiometric coefficients used in the chemical equation as exponents of the reactants or products. To demonstrate using an equation that had A + 2B = 4C and assuming they are all aqueous solutions, the Kc would be calculated through [C]^4/[A][B]^2. However if you halved all the stoichiometric coefficients Kc would now equal [C]^2/[A]^1/2[B], which would obviously change the Kc value even if the concentrations of each reactant or product has not changed.

### Re: Changing Kc

Posted: **Sat Dec 09, 2017 4:24 pm**

by **Nora Sharp 1C**

You are correct that K for a reaction only changes when temperature changes. However, when we halve all the stoichiometric coefficients for the reaction, we are actually changing the reaction we are using in our calculations.

2SO2 + O2 ---> 2SO3 is different than SO2 + (1/2)O2 ---> SO3.

Therefore, the K value is different for both reactions. You may argue that the two reactions above have the same reactants and same products, and that's true, but mathematically, we have to change the K value to keep our results accurate. That's why when we're changing K for the reactions at 400K, the answer x^(1/2) only works if both reactions are at 400K, because in this case, only the stoichiometric coefficients are changing. In this scenario, when 400K is constant, K changes for each equation only according to how to coefficients change.