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For the reaction: `X(g) rarr Y(g)+Z(g)`, the following data were obtained at `30^(@)C`: `{:("Experiment",[X](mol L^(-1)),"Rate" (mol L^(-1) hr^(-1))),(I,0.17,0.05),(II,0.34,0.10),(III,0.68,0.20):}` The equilibrium constant for the reaction is `0.50`. Assuming that the reaction proceeds by one-step mechanism. Find the rate constant of reverse reaction?A. `0.294 hr^(-1)`B. `0.588 hr^(-1)`C. `0.123 hr^(-1)`D. `0.117 hr^(-1)` |
Answer» Correct Answer - B `X(g) underset(k_(b))overset(k_(f))hArr Y(g) + Z(g)` `k_(eq)=(k_(f))/(k_(b))` `impliesK_(b)=(0.294hr^(-1))/(0.50)= 0.588hr^(-1)` |
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