For a chemical reaction, the plot of [ R ] versus time t yields a straight line with a slope of -0.02 s ⁻¹ . How much time will it take for the concentration of the reactant to drop from 0.8 mol L ⁻¹ to 0.1 mol L ⁻¹ ? (Given : 2 = 0.693 , 2 = 0.301 )

Step-by-step solution

For a first-order reaction, the integrated rate equation is: [ R ] = -kt + [ R ]₀ The plot of [ R ] versus t is a straight line with slope = -k . Given slope = -0.02 s ⁻¹ , we have k = 0.02 s ⁻¹ . The half-life of the reaction is: t_ 1/2 = 2 k = 0.693 0.02 = 34.65 s The concentration drops from 0.8 mol L ⁻¹ to 0.1 mol L ⁻¹ , which is a reduction to 1 8 of the initial amount. This corresponds to 3 half-lives ( 2^3 = 8 ). Total time t = 3 t_ 1/2 = 3 34.65 = 103.95 s . Using the zero-order formula gives 35.0 s , and m