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How Fast? The Rate of Chemical Change · Paper 2 — Free Chemistry HL Practice Questions

1ChallengeSAQ-LRate law and order of reactions9 marksPaper 2~14 min

Data

t / st\,/\,\text{s} — 00 — 120120 — 240240 — 360360 — 480480 — 600600 [OH−] / mol dm−3[\text{OH}^-]\,/\,\text{mol}\,\text{dm}^{-3} — 0.1000.100 — 0.0750.075 — 0.0560.056 — 0.0420.042 — 0.0320.032 — 0.0240.024
Ethyl methanoate (HCOOC2H5\text{HCOOC}_2\text{H}_5) undergoes hydrolysis in alkaline solution: HCOOC2H5(aq)+OH−(aq)→HCOO−(aq)+C2H5OH(aq)\text{HCOOC}_2\text{H}_5\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{HCOO}^-\text{(aq)} + \text{C}_2\text{H}_5\text{OH(aq)} A student investigates the kinetics at 25 ∘C25\,^\circ\text{C} with initial concentrations [ester]0=[OH−]0=0.100 mol dm−3[\text{ester}]_0 = [\text{OH}^-]_0 = 0.100\,\text{mol}\,\text{dm}^{-3}. Samples are withdrawn and titrated with standard acid to give the following
(a)
Using the data above, show that the reaction is second order with respect to OH−\text{OH}^-. [2 marks]
(b)
Calculate the rate constant kk, including its units. [2 marks]
(c)
Explain why the order with respect to the ester cannot be determined from this data alone. [2 marks]
(d)
A student proposes repeating the experiment using [ester]0=0.100 mol dm−3[\text{ester}]_0 = 0.100\,\text{mol}\,\text{dm}^{-3} and [OH−]0=1.00 mol dm−3[\text{OH}^-]_0 = 1.00\,\text{mol}\,\text{dm}^{-3}. Deduce how this modification allows the order with respect to the ester to be determined, and identify what quantity should be plotted to confirm a first-order dependence on the ester. [3 marks]
diagram

Solutions

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