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The rate constant of a reaction with respect to the reactant A is `6 min^(-1)`. If we start with [A] = `0.8 mol L^(-1)`, when would [A] reach the value of `0.08 mol^(-1)`? |
Answer» For the first order reaction, `t=2.303/k log a/(a-x)` `a=0.8 mol L^(-1),(a-x)=0.008 mol L^(-1), k=6min^(-1)` `t = (2.303)/(6min^(-1)) log (0.8 mol L^(-1))/(0.08 mol L^(-1)) = 2.303 / (6 min^(-1)) log 10 = 0.38 min` |
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