sampling question 5
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sampling question 5
Does anyone know how to do sampling question 5? It's the question that gives us multiple Kc and reaction equations and asks us to solve for the overall K
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Re: sampling question 5
Hi! Basically you have to mix and match the equations to make the overall reaction you are trying to solve for! Sometimes this would include multiplying a whole reaction by a certain coefficient, or even reversing the reaction. When you make these changes to the reactions, you have to also modify their K values. So for example, if you multiply a whole reaction by a coefficient of two, you have to square the K value of that reaction. And when you reverse a reaction, you must take the inverse of the K value of that reaction, so K^-1.
Once you have the modified K values for the reactions that you matched up, you multiply those two K values together to get the K value for the overall reaction.
Hope this helps!
Once you have the modified K values for the reactions that you matched up, you multiply those two K values together to get the K value for the overall reaction.
Hope this helps!
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Re: sampling question 5
You have to sum two reactions to obtain the overall reaction and find the equilibrium constant.
When you add two chemical reactions together, the equilibrium constants are multiplied.
When you reverse a chemical reaction, the equilibrium constant is inversed (^-1).
If you multiply a chemical reaction by a coefficient A, the equilibrium constant is raised to the power of A (K^A).
Hope this helps!
When you add two chemical reactions together, the equilibrium constants are multiplied.
When you reverse a chemical reaction, the equilibrium constant is inversed (^-1).
If you multiply a chemical reaction by a coefficient A, the equilibrium constant is raised to the power of A (K^A).
Hope this helps!
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Re: sampling question 5
Break down the main reaction into its smaller parts and you will find that the other reactions given will make more sense!
Also remember that when a reaction is multiplied by a coefficient n, the Ka will change to Ka^n.
When two reactions are added together, the Ka values are multiplied.
I hope this helps! :)
Also remember that when a reaction is multiplied by a coefficient n, the Ka will change to Ka^n.
When two reactions are added together, the Ka values are multiplied.
I hope this helps! :)
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