Three gaseous reactions are at simultaneous equilibrium in a closed vessel. (equilibrium constant ) — Chemical Equilibrium Chemistry Question
Question
Three gaseous reactions are at simultaneous equilibrium in a closed vessel. $A + 2B \rightleftharpoons C$ (equilibrium constant $K_1 = 10$) $C + D \rightleftharpoons E$ (equilibrium constant $K_2 = 2$) What is the numerical value of the equilibrium constant for the net reaction $A + 2B + D \rightleftharpoons E$?
Answer: 20
💡 Solution & Explanation
The net reaction $A + 2B + D \rightleftharpoons E$ is obtained by directly adding the two given step reactions: $(A + 2B \rightleftharpoons C) + (C + D \rightleftharpoons E)$. When chemical reactions are added, their equilibrium constants must be multiplied together. Therefore, $K_{net} = K_1 \times K_2 = 10 \times 2 = 20$.
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