Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 17 - Chemical Equilibrium - Exercises - Calculation of K - Page 702: 24



Work Step by Step

Here $K_{c}$=$\frac{[NO]^{2}}{[N_{2}][O_{2}]}$ Given [NO]=$1.1×10^{-5} mol/L $ $[N_{2}]=6.4×10^{-3} mol/L$ And $[O_{2}]=1.7×10^{-3} mol/L$ Then $ K_{c}=\frac{[1.1×10^{-5}]^{2}}{[6.4×10^{-3}] [1.7×10^{-3}]}$ =1.1121×$10^{-5}$
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