- Questions

# 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!