For A+2B <-> 2C and Kc = 2.13

For 2C <-> D and Kc = 0.254

Calculate the value of equilibrium contant for the reaction D <-> A+2B

I'm so sorry to bother anyone.. but am i only supposed to multiple 2.13 and 0.254 together in order to get the third value of Kc? That's the vibe i'm getting from my notes.. but that seems too simple so i'm scared it's wrong

2 answers

A + 2B ==>2C k1 = 2.13
2C ==> D k2 = 0.254
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A + 2B ==> D k3 = k1*k2
But you want the reverse of that reaction. You want
D ==> A + 2B so k4 = 1/k3
Thank you so much Dr. Bob
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