IB Chemistry R2.2: How fast? The rate of chemical change worked solutions

Subtract readings and times separately: (0.500−0.200)/(90.0−30.0)=0.00500 mol dm−3 s−1. Option B [1]. Use the change in product concentration, rather than its final value, and divide by the interval between the two readings.

At fixed temperature the reacting number densities double on compression, so encounters become more frequent. The normalised energy distribution and the fraction above the unchanged barrier stay fixed. Option D [1].

The alternative route has smaller barriers in both directions. Its endpoints are the original species, so their enthalpy difference is unchanged. Option A [1].

The concentration ratio is 0.120/0.040=3 and rate ratio is (1.8×10−4)/(2.0×10−5)=9. Solve 3^m=9: m=2. Option C [1].

The valley between successive maxima represents an intermediate with a finite lifetime. A barrier maximum represents a transition state. Option B [1].

Subtract the two linear Arrhenius expressions: ln(k2/k1)=(50000/8.31)(1/300−1/310)=0.646972814… . Exponentiate: k2/k1=1.909751…=1.91. Option D [1]. Subtract the reciprocal absolute temperatures with the lower temperature first, so the logarithmic ratio is positive. Exponentiate to obtain the increase factor.

B [1]. Calibrated absorbance at a suitable wavelength tracks the coloured reactant concentration. The other proposed signals do not measure its removal in this experiment.

A line through the origin means r\propto[Q], so the order is one [1]. Tripling [Q] therefore triples the rate [1].

The prefactor ratio is 0.20/0.05=4 [1]. Equal activation energies and temperature make the exponential factors equal, so the rate-constant ratio is also four [1].

The most probable energy rises, moving the peak right [1]. The distribution spreads over a wider energy range with the same total area, so the peak is lower [1].

(a) Specify a species and the change in its concentration per elapsed time [1]. (b) (0.0840−0.0300)/90.0 [1]=6.00×10−4 mol dm−3 s−1 [1]. (c) The changing gradient means that an interval average need not equal the tangent at the start [1]. (d) Draw labelled concentration/time axes, a trace from zero rising ever less steeply into a plateau [1], and final marker concentration 2×0.0840=0.168 mol dm−3 [1]. At 90.0 s the marker concentration is 2(0.0840−0.0300)=0.108 mol dm−3; this is a key consistency check, not an extra printed demand.

After complete dissolution, the original crystals are absent. Equal concentrations, temperature, portions and remaining conditions make the reacting liquids equivalent; the solid particle-size comparison does not survive dissolution. Option B [1].

The MCQ answer is B: both solutions have the same dissolved composition. For the separate solid test, total area at fixed volume is 6V/a, so S_{1.5}/S_{4.5}=4.5/1.5=3 [1]. The smaller cubes expose more reactive sites [1], allowing more reagent encounters at the solid surface per second at the same temperature [1]. No exact rate multiplier follows without a kinetic model.

Set the reactant level to zero. The transition state is +73 and the products +28 kJ mol−1. The reverse barrier is 73−28=45 kJ mol−1. Option D [1].

The original reverse barrier is 73-28=45\,\mathrm{kJ\,mol^{-1}}, option D [1]. For the second path the reverse barrier is 51-28=23\,\mathrm{kJ\,mol^{-1}} [1]. The same endpoints give \Delta H=+28\,\mathrm{kJ\,mol^{-1}} [1].

The total concentration power is three. Units(k)=(mol dm−3 s−1)/(mol dm−3)^3=(mol dm−3)^−2 s−1=dm6 mol−2 s−1. Option B [1].

Overall order is three, so the MCQ answer is B [1]. Substitute into the supplied law: r=2.50(0.0800)(0.0400)^2=3.20\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}} [2]. Halving N multiplies its squared factor by (1/2)^2=0.250 [1].

(a) Cancel CL and CL2 and the carrier: net 2L+N→L2N [1]; C is returned [1]. (b) Elementary slow-step law r=k2[CL][L] [1]; substitute [CL]=K[C][L], giving r=k2K[C][L]^2=kobs[L]^2 when free C is fixed [1]. (c) Doubling L multiplies rate by four [1]; doubling N leaves it unchanged within this model [1]. (d) CL and L are two colliding particles. The additional L dependence comes from the prior equilibrium; the slow step is bimolecular [1].

(a) x=1000 K/T gives standard gradient −5400 K [1]; Ea=5400×8.31=44874 J mol−1=44.9 kJ mol−1 [1]. (b) Intercept is ln(A/(1 s−1))=16.2 [1]; A=exp(16.2)=1.08535…×10^7 s−1, printed 1.09×10^7 s−1 [1]. (c) At 300 K, x=10/3 and y=−1.80 [1]; k=exp(−1.80)=0.165 s−1 [1]. (d) Under the stated fixed-A assumption, the catalyst line shares the intercept and has a shallower negative slope [1]. The fixed-A comparison is stipulated; other pathways may have different prefactors.

Runs 1 and 2 double only X and double the rate, so the X order is one [1]. Runs 1 and 3 double only Y and quadruple the rate, so its order is two [1]. Thus r=k[X][Y]^2 [1]. From run 1, k=(2.00\times10^{-4})/(0.0500\times0.100^2)=0.400\,\mathrm{dm^6\,mol^{-2}\,s^{-1}} [2]. The new rate is 0.400(0.0750)(0.0500)^2=7.50\times10^{-5}\,\mathrm{mol\,dm^{-3}\,s^{-1}} [1].

(a) Label energy horizontally and number density per energy interval vertically [1]. Draw two non-negative skewed curves with the hotter peak lower and farther right, preserving total area [1]. Mark the same activation threshold for both [1]. (b) A larger fraction has energy at least E_a, increasing the number of sufficiently energetic encounters per time [1]. (c) A catalyst supplies a route with a lower activation threshold; the distribution at fixed temperature is unchanged [1]. The accompanying solution figure shows the equal-area curves and common threshold; the lower catalytic threshold leaves the fixed-temperature curve unchanged.

(a) Replace varying stock-Q volumes with water to preserve a fixed Q-portion volume [1]; mix each with the same other-reactant stock volume/concentration, preserving final volume [1]. (b) Thermostat: temperature changes the energetic fraction [1]. Keep catalyst loading fixed: it can independently alter the rate [1]. A well-explained consistent mixing/start protocol is acceptable instead. (c) Record absorbance promptly at repeated short intervals and apply the supplied calibration [1]; plot product concentration/time and obtain initial tangents [1]; repeat each Q setting and compare mean initial gradients with repeat spread [1].

(a) Label energy/progress, reactant 0 and product −55 [1]; maxima +48 and +40 [1]; intermediate valley −22 and transition-state peak labels [1]. (b) Ea1=48−0=48 and Ea2=40−(−22)=62 kJ mol−1 [1]. The stated comparable-factor/local-barrier criterion selects step 2 [1]. (c) An intermediate is a species occupying a minimum, formed before a later consumption step [1]. A transition state is the configuration at a barrier maximum [1].

(a) Fixed increment/time gives 0.00120/100=1.20×10−5 and 0.00120/25.0=4.80×10−5 mol dm−3 s−1 [1]. Divide each by 0.120: k295=1.00×10−4 s−1 [1], k315=4.00×10−4 s−1 [1]. (b) Ea=8.31 ln(4)/(1/295−1/315) [1]=53525.293… J mol−1=53.5 kJ mol−1 [1]. (c) Plot ln(k/(1 s−1)) vertically against 1/T horizontally [1], gradient −Ea/R [1]. (d) Two endpoint means cannot test linearity; extra temperatures and repeats are needed. Fixed-increment average rates also approximate initial rates only under negligible depletion [1 for one explained limitation]. Connection label R2.2 + Tools + Inquiry.

(a) Divide the Arrhenius expressions: k_c/k_u=0.00100\exp(15000/(8.31T)) [1]. At 300 K this is 0.410 [1]. (b) The stipulated parallel-path sum gives k_{\rm total}/k_u=1+0.410=1.41 [1]. The original route remains available, and the extra route makes an additional positive contribution [1]. (c) Set the ratio to one: 15000/(8.31T)=\ln1000 [1], giving T=261\,\mathrm K [1]. (d) The much smaller prefactor offsets some of the lower barrier advantage [1]. Both paths connect the same reactants and products, so \Delta H is unchanged [1].

(a) 36.0/60.0=0.600\,\mathrm{cm^3\,s^{-1}} [1]. Use P=100000\,\mathrm{Pa} and V=36.0\times10^{-6}\,\mathrm{m^3} [1]: n_G=PV/(RT)=1.45373\times10^{-3}\,\mathrm{mol}, or 1.45\times10^{-3}\,\mathrm{mol} [1]. (b) Q loss is twice the gas amount [1]. Divide by the liquid volume and elapsed time [1]: 2n_G/(0.0500\times60.0)=9.69\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}} [1]. Keep guard digits from the gas calculation. (c) Lost gas makes the inferred disappearance rate too small [1]. Check the assembled seal using a known gas volume or a pressure-hold test before reacting [1].

(a) \Delta H=-27-15=-42\,\mathrm{kJ\,mol^{-1}} [1]. Forward barriers are 81-15=66 [1] and 54-15=39\,\mathrm{kJ\,mol^{-1}} [1]. (b) Catalytic reverse barrier is 54-(-27)=81\,\mathrm{kJ\,mol^{-1}} [1]. (c) Released heat has magnitude 0.0100\times42000=420\,\mathrm J [1]. With all heat absorbed by the stated liquid, \Delta T=420/(80.0\times4.18)=1.26\,\mathrm K [1]. (d) The catalyst offers a lower-barrier pathway [1], while self-heating increases the fraction of particles above the threshold; temperature must be controlled to separate these effects [1].

(a) Cancel I to obtain X+2Y\rightarrow XY_2 [1]. The slow elementary event gives r=k_2[I][Y] [1]; substituting the given equilibrium relation yields r=k_2K_c[X][Y]^2 [1]. (b) [I]=6.00(0.0400)(0.0200)=0.00480\,\mathrm{mol\,dm^{-3}} [1]. Then r=0.250(0.00480)(0.0200)=2.40\times10^{-5}\,\mathrm{mol\,dm^{-3}\,s^{-1}} [1]. (c) The factor is 2\times2^2=8 [1]. (d) The fast step has approximately balanced forward/reverse fluxes on the slower timescale [1], while the slow step continues removing I and accumulating product; the entire system has no stipulated balance of overall opposing rates [1].