CHEM 2 Question

For A+2B <---> 2C : K=2.23
For 2C <---> D : K=0.27
What is the value of the equilibrium constant for the reaction: 2D <---> 2A + 4B ?

I calculated it and got this:
k=(k1)(k2)
=(2.23)(.27)=0.6021
=(1/0.6021)=1.67
=(1.67)^2= 2.76

Is this answer correct? I am unsure whether or not I am supposed to square the 1.67. I squared it because the value of A and B was doubled, so is that correct? Please help! Thanks!

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  1. Yes, you did it correctly.
    http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Equilibrium_Constants_From_Other_Constants.htm

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    bobpursley

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