IB Chemistry R2.3: How far? The extent of chemical change worked solutions

Conversions persist in both directions. Equal opposing rates keep each bulk concentration steady, so A fits; neither equality of concentrations nor a stopped process follows.

The first repeated pair is the pair at 20 s: the 30 s reading matches it [1]. Each species has its own steady concentration, which supports equilibrium within the measurement resolution [1]. The data locate no more precise onset between 10 s and 20 s. Equality of the two species concentrations is unnecessary.

Invert 0.0040 to obtain 250 for the reversed equation. The physical sample is unchanged when its equation is reversed. X remains the more abundant form; choose C.

Halving every coefficient takes the square root, so K(\mathrm P\rightleftharpoons\mathrm R)=\sqrt{49.0}=7.00 [1]. Reversing this one-to-one equation gives K=1/7.00=0.142857\ldots=0.143 [1]. Raising a ratio to a power follows directly from changing all concentration exponents.

Use the measured equilibrium values: K=(0.150)^2/0.0500=0.450. The coefficient of the atomic species becomes an exponent, giving B.

Compare identical starting mixtures at fixed temperature. The catalyst changes the time needed for equilibration. It supplies no new equilibrium ratio, so both final compositions coincide; B.

The violet observation means the new equilibrium contains a larger product-to-reactant ratio. Heating favours the heat-absorbing direction, so the forward conversion is endothermic [1]. For this one-to-one model, the increase in that ratio is an increase in K [1]. A faster reaction alone would not explain a changed final colour.

Lowering temperature favours the exothermic forward conversion [1]. Isothermal compression favours the side with one mole of gas per reaction event rather than two, increasing the equilibrium amount of C [1]. The temperature change changes K; compression at the original temperature does not.

Evaluate methanol concentration divided by carbon monoxide concentration and the square of hydrogen concentration. The four results are 100, 250, 500 and 25. Only D gives the supplied equilibrium value.

Convert the standard energy to -5000 J mol−1. Then ln K=5000/(8.31×400)=1.50421 and K=exp[5000/(8.31×400)]=4.50060. Choose C.

(a) Molecules leave the liquid [1] while other molecules return at the same rate [1]. (b) More limonene enters the vapour at the new equilibrium [1]; the supplied endothermic sign makes vaporisation favoured on warming [1]. Liquid must remain.

(a) Multiply the two product factors and divide by the two reactant factors, retaining powers 2 and 4 [1]. (b) K=[X₂]²[Y₂]/[X₂Y]² [1]. (c) Every component is gaseous, so there is one phase [1]; each balanced coefficient supplies its concentration power [1].

(a) Forming 0.120 mol E consumes 0.120/2=0.0600 mol D. Thus n(\mathrm D)=0.300-0.0600=0.240 mol [1]. (b) Divide by 2.00\,\mathrm{dm^3}: [\mathrm D]=0.120 and [\mathrm E]=0.0600\,\mathrm{mol\,dm^{-3}} [1]. (c) K=[\mathrm E]^2/[\mathrm D]=0.0600^2/0.120=0.0300 [1]. The volume factors leave a net 1/V: the unconverted amount ratio is 0.120^2/0.240=0.0600, twice the correct numerical concentration ratio [1].

(a) Thermostat the sealed cell because K depends on temperature [1]. Wait for repeated readings to be stable within the probe precision; stronger evidence is the same endpoint approached from P-rich and R-rich mixtures with the same total concentration [1]. Maintaining the seal and checking calibration are also relevant controls; award at most two marks for distinct justified checks. (b) Material balance gives [\mathrm P]=0.180-0.120=0.0600\,\mathrm{mol\,dm^{-3}} [1]. Hence K=[\mathrm R]/[\mathrm P]=0.120/0.0600=2.00 [1].

(a) Extra dissolved sulfate favours precipitation [1], consuming some sulfate with barium ions [1]. (b) The ion concentrations at equilibrium stay the same [1], since a pure solid phase is still present at the same temperature [1]. (c) More BaSO₄ dissolves [1], supplying some of the barium ions removed [1].

(a) Start with horizontal traces at the stated original values. At 40 s, draw J vertically from 0.100 to 0.160 [1]. L is continuous through 40 s; there is no instantaneous dilution or addition of L [1]. Draw a smooth J decline to 0.120 and an L rise to 0.240, flattening into steady new plateaux [1]. (b) The net forward reaction consumes only 0.040 of the added 0.060\,\mathrm{mol\,dm^{-3}} of J, leaving an excess of 0.020 over its original concentration [1]. The response does not undo the entire pulse. The supplied endpoints conserve [\mathrm J]+[\mathrm L] after addition and preserve the ratio 2.

(a) For 2X⇌X₂ write Q=[X₂]/[X]². Expansion halves both concentrations [1], so Qnew=K×(1/2)/(1/2)²=2K [1]. (b) Q exceeds K [1]; X₂ is consumed to make X [1]. (c) Heating the rigid cell leaves instantaneous concentrations unchanged; its Q is the old K, whereas the exothermic forward reaction has a smaller new K [1]. Net dissociation follows [1].

(a) ln K=5400/(8.31×360)=1.80505 [1]; exponentiating gives 6.08030, or 6.08 [1]. (b) The reversed ratio gives 0.164466, or 0.164 [1]; reversing the energy account gives +5.40 kJ mol−1 [1]. (c) The actual reaction energy includes RT ln Q [1]. With Q=K this term cancels ΔG°, so the actual value is zero [1].

(a) Set the remaining parent concentration approximately equal to 0.250 [1]. Equal product changes give K≈x²/0.250 [1], hence x=sqrt(4.00×10−8×0.250)=1.00×10−4 mol dm−3 [1]. (b) 100x/0.250=0.0400% [1]; the neglected decrease is small enough for this estimate [1]. (c) The fractional change sqrt(K/c₀) grows as c₀ falls, so a small K alone is insufficient [1].

(a) Added H⁺ drives protonation of HS⁻ [1], producing more dissolved H₂S [1]. (b) This supplies extra dissolved material to the second equilibrium [1]; transfer to its gas phase increases [1]. (c) Neither temperature nor either reaction equation changes, so both K values stay fixed [1]. Composition changes to satisfy those same conditions after the added reagent [1].

(a) At 4 bar, heating reduces conversion from 86.0% to 61.0% because it favours the endothermic reverse direction [1]. At 420 K, compression favours one gas mole of C over two gas moles of reactants, explaining the rise to 76.2% [1]. (b) The hotter high-pressure process has faster stated production and a higher conversion than the hotter low-pressure trial [1]. Against the cooler trial it sacrifices 9.8 percentage points of equilibrium conversion and adds compression/equipment costs; profit or energy data are needed for a final choice [1]. (c) A catalyst shortens the time to the same equilibrium at the chosen temperature [1]. Separating product and recycling unused reactants allows further conversion on later passes; it improves overall feed use without increasing the equilibrium constant. Neither catalyst nor recycle changes K at fixed temperature [1].

(a) ln K=−17400/(8.31×298) [1]; K=0.000888136, or 8.88×10−4 [1]. (b) Insert K=0.130 and T=1000 K into −RT ln K [1]; obtain +16.9542 kJ mol−1, or +17.0 [1]. (c) K=1 sets ΔG°=0, so T=ΔH°/ΔS° [1]; 57000/176=323.864 K, or 324 K [1]. (d) ΔG°=57000−500×176=−31000 J mol−1 [1]. Then ln K=31000/(8.31×500) [1]; K=1738.70, or 1.74×10³ [1].

(a) Ammonia accepts a proton to give the conjugate acid \mathrm{NH_4^+} [1]. Hydronium donates a proton to leave its conjugate base \mathrm{H_2O} [1]. (b) Acid addition promotes protonation, lowering the freely dissolved ammonia relative to the undisturbed mixture [1]. The gas/solution equilibrium then transfers more ammonia into water; less gaseous ammonia gives a lower ammonia partial pressure at fixed headspace volume and temperature [1]. (c) Neither constant changes because temperature stays fixed [1]. At the final equilibrium, ammonia still enters and leaves the water at equal rates; steady pressure is compatible with molecular transfer [1].

(a) Q=0.0800^2/0.0400=0.160 [1]. Since Q<0.200, net dissociation increases the product ratio [1]. (b) n=(0.0400+0.0800)\times2.00=0.240\,\mathrm{mol} [1]. Convert V=2.00\times10^{-3}\,\mathrm{m^3}; p=nRT/V=398880\,\mathrm{Pa}=399\,\mathrm{kPa} to three significant figures [1]. (c) From an equilibrium mixture, both concentrations double. Thus Q_{\rm new}=(2[\mathrm A])^2/(2[\mathrm{A_2}])=2K=0.400 [2]. (d) Q>K drives net reverse conversion, reducing the quotient [1]. Two gas particles combine into one dimer, so the direction also agrees with favouring fewer gas particles under compression [1]. The later pressure cannot be found by simply doubling the old equilibrium pressure because reaction changes the total amount of gas.

(a) \Delta G^\circ=-RT\ln K=-8.31\times350\ln9.00=-6390.63\,\mathrm{J\,mol^{-1}} [1], or -6.39\,\mathrm{kJ\,mol^{-1}} [1]. (b) Heating lowers the equilibrium product/reactant ratio from 9.00 to 3.00, so it favours the reverse direction [1]. Heating favours an endothermic direction; therefore the forward direction is exothermic [1]. (c) The shorter time with the same endpoint shows faster approach to equilibrium, through acceleration of both opposing processes, with unchanged K at 350 K [1]. At equilibrium the actual reaction Gibbs energy change is zero, even though the standard value is negative [1]. The time ratio alone does not determine an activation energy or a rate law.