Suggested Solution: Tutorial 6 Reaction Kinetics Self CheckQuestion 1NoTermDefinition(a)Rate equation/lawThe rate equation or rate law is a mathematical equation that shows how the rate of reaction is dependent on the concentrations of the reactants; it relates the rate of the reaction to the concentration of reactants raised to the appropriate power(b)Rate constantThe rate constant, k is a proportionality constant in the rate equation of the reaction(c)Order of ReactionThe order of reaction with respect to a reactantis the power to which the concentration of that reactant is raised to in the rate equation.The overall order of reactionis the sum of the powers to which the concentrations of the reactants are raised to in the rate equation(d)Half-lifeThe time required for the concentration of a reactant to decrease to half of its initial concentrationQuestion 2Rate equationrate = k[H2O2]rate = k[H2][O2]rate = k[NO2]2Rate = kUnit of ks-1mol-1 dm3s-1mol-1 dm3s-1mol dm-3 s-1Instead of s, min may also be used.Question 3Order of rxn(i) Graph of [reactant] against time(ii) Graph of [product] against time(iii) Graph of rate against [reactant]ZeroFirstSecond1[rxt]/ mol dm-3Time/sC0½C0¼C0[pdt]/ mol dm-3Time/sC∞½C∞¾C∞[rxt]/ mol dm-3Time/sC0½C0¼C0[rxt]/ mol dm-3Time/sC0½C0¼C0[pdt]/ mol dm-3Time/sC∞½C∞¾C∞[pdt]/ mol dm-3Time/sC∞½C∞¾C∞rate/ mol dm-3 s-1[rxt]/ mol dm-3rate/ mol dm-3 s-1[rxt]/ mol dm-3rate/ mol dm-3 s-1[rxt]/ mol dm-3

Question 4(a)Half-life (t1/2) of the reaction is constant.(b)Draw the tangent to the graph at point P (as shown in Figure 1a).

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(c)Draw tangents to the graph at 5 or more different [reactant]. Find the gradients as described in (b)

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Standard QuestionsQuestion 1(a)Comparing results of experiments 1 and 2, when [Y] increases 3 times, rate of increase in [XY2] increases by 9 times ⇒order of reaction with respect to Y is 2.Comparing results of experiments 1 and 5, when [X] increases 3 times, rate of increase in [XY2] remains the same ⇒order of reaction with respect to X is 0.(b)Rate = k[Y]2Overall order of reaction = 2(c)Using results from experiment 1, 0.0001 = k(0.10)2⇒k = 0.01 mol-1dm3s-1(d)Using value of k calculated from (c), rate = 0.01[Y]2For experiment 6, 0.0036 = 0.01a2⇒a = 0.00360.01=0.60For experiment 3, b = 0.01(0.30)2= 0.0009(e)Hence, mechanism 1is consistent with the observed kinetics.Question 2 Rate = k[A]2[B], original rate = R= k[A]02[B]0 where [X]0 = initial [X]Qn.(a)(b)(c)(d)(e)ChangeDouble partial pressures of A & BPartial pressure of A doubled.Double volume of reacting vessel.Double total pressure with inert gas.30°C increase in temperature.New rate8R4R⅛RR8RReasonDoubling partial pressure ≡doubling concentration