ChemistryHardHigh School (11–12)Chemical Kinetics › Integrated Rate Equations and Half-LifeJEE AdvancedJEE Main
Created by StemPy LabMethyl acetate is hydrolysed in dilute hydrochloric acid at : . Water is in large excess and the acid is a catalyst, so the reaction is pseudo first order in the ester. At different times, a portion of the mixture is taken out, cooled in ice and titrated with the same sodium hydroxide solution. The titres are at the start, after , and once the reaction is complete. Take . What is the half-life of the ester, in , to the nearest integer?
← All Quizzes
← A first order reaction in solution is run twice, from the same starting concentration, and followed by measuring the concentration of the reactant. At 300 K, it takes 36.0 min for 25 per cent of the reactant to be used up. At 320 K, the same 25 per cent is used up in only 12.0 min. Assume that the activation energy and the pre-exponential factor do not change between the two temperatures, and take R = 8.314 J K⁻¹ mol⁻¹. What is the activation energy of the reaction, in kJ mol⁻¹, to the nearest integer?Many enzyme reactions follow the rate law -(d[S])/dt = (Vmax[S])/(Km + [S]), where S is the substance the enzyme acts on. This rate is nearly zero order when [S] is much larger than Km, and nearly first order when [S] is much smaller. For one enzyme at a fixed temperature, Vmax = 1.00 × 10⁻³ mol L⁻¹ min⁻¹ and Km = 5.00 × 10⁻³ mol L⁻¹. The amount of enzyme stays constant. A solution starts with [S]₀ = 2.00 × 10⁻² mol L⁻¹. How long does it take for [S] to fall to 5.00 × 10⁻³ mol L⁻¹? Give the answer in min, to two decimal places. Take ln 2 = 0.6931. →
Discussion (0)
Log in to join the discussion.
Loading…