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The progress of the reaction AiffnB , with time is presented in the figure. Determine (i) the value of n (ii) the equilibrium constant K (iii) the initial rate of conversion of A |
Answer» SOLUTION : At equilibrium Rate (forward) = Rate (backward) ` = (0.6 - 0.3)/(5)= (0.6)/(5) XX 1/n` For the eqb: `A IFF 2B, K =(0.6^2)/(0.3)` Initial rate = change in concentration of A in the 1ST hour 2,1.2,0.1 |
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