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For the reaction<br>(i) $$A \rightleftharpoons B$$; $$K_c = 2$$<br>(ii) $$B \rightleftharpoons C$$; $$K_c = 4$$<br>(iii) $$C \rightleftharpoons D$$; $$K_c = 6$$<br>For the reaction $$A \rightleftharpoons D$$ what is the $$K_c$$?
Correct Answer: 48
Adding the three reactions gives $$A \rightleftharpoons D$$, and when reactions are added their equilibrium constants multiply. So $$K_c = 2 \times 4 \times 6 = 48$$.
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