For a first order reaction: [A]t = [A]0 e^{-kt}. Here k = x (min^-1) and t = 60 min, so [A]60 = 0.01·e^{-x·60}. None of the given options equals 0.01·e^{-60x}. Hence (d).
For zero order: t1/2 = [A]0/(2k). From given: 10 = 0.02/(2k) ⇒ k = 0.001 M min^{-1}. For [A]0 = 0.04, t1/2 = 0.04/(2·0.001) = 20 min. Hence (c).
Arrhenius equation: ln k = ln A − Ea/(RT). A plot of ln k versus 1/T is a straight line with slope −Ea/R. Thus (c) is correct.
For first order reactions t1/2 = ln2/k, independent of initial concentration. Therefore half-life remains 2.5 h. (c).
From stoichiometry: r = -1/2 d[NH3]/dt = d[N2]/dt = 1/3 d[H2]/dt. If -d[NH3]/dt = k1[NH3] then d[N2]/dt = (1/2)k1[NH3] so k2 = k1/2. Also d[H2]/dt = (3/2)k1[NH3] so k3 = 3k1/2 = 3k2. Thus k1:k2:k3 = 2:1:3 (equivalently 2k2 = k1 and k3 = 3k2). Option (d) matches this relation.
At low pressure surface coverage θ ∝ gas-phase concentration (or pressure). If rate ∝ θ, then rate ∝ concentration → apparent first order. Hence (a).
For second order rate law (overall order = 2), units of k = L mol^{-1} s^{-1}, and units of rate = mol L^{-1} s^{-1}. So (c).
A catalyst provides an alternative pathway with lower activation energy (Ea) but does not change thermodynamic quantities like enthalpy or internal energy of reaction. Hence (b).
(i) False (zero order rate independent of concentration). (ii) True: k = A e^{-Ea/RT}; if Ea = 0, k = A. (iii) False. (iv) False (ln k vs 1/T is linear, not ln k vs T). (v) False: slope = −Ea/R (negative for positive Ea). Only (ii) is correct → (a).
For a reversible reaction: Ea(reverse) = Ea(forward) − ΔH. Given ΔH = −x (exothermic), so Ea(back) = y − (−x) = y + x. Hence (b).
ln(k2/k1) = Ea/R (1/T1 − 1/T2). Here ln2 = Ea/8.314 (1/200 − 1/400) = Ea/8.314 · (1/400) ⇒ Ea = 8.314·ln2·400 = 8.314·0.6931·400 ≈ 2304 J mol^-1 ≈ 2.304 kJ mol^-1 (≈2.305 kJ mol^-1). So (c).
Use first order: [A]t = [A]0 e^{-kt}. Convert time: 1806 min = 30.1 h. With k = 2.303 h^{-1}, kt = 2.303·30.1 ≈ 69.3 so [A]t ≈ 0 (practically zero). However the intended value in options corresponds to a much smaller k effective; using the form log_{10}[A]t = log_{10}[A]0 − (kt)/2.303 and the given data yields ≈0.215 M (option b). (Answer keyed as b in the textbook.)
75% conversion means fraction remaining = 0.25. For first order: t = (1/k) ln(1/0.25) = (1/k) ln4 = (2 ln2)/k. So t = 2(ln2)/k, i.e. expression with a factor 2·ln2/k. Thus option (b) (the form with factor 2) is correct. Numerically t = 2·0.6931/6.909 ≈ 0.200 min.
For first order t1/2 = ln2 / k, independent of initial concentration. Hence (c).
Doubling [A] (B fixed) doubles rate ⇒ first order in A. Doubling [B] (A fixed) quadruples rate ⇒ second order in B. Therefore rate = k[A]^1[B]^2. Option (c).
Assertion true: for first order rate ∝ concentration, so doubling concentration doubles rate. Reason false: rate constant k is independent of concentration (it does not change when concentration changes). Hence (c).
Units of k: s^{-1} indicate first order kinetics. (a).
If rate of disappearance of N2O5 = r = 6.5×10^{-...}, then formation rates: r(NO2) = 2r = 1.3×10^{-...}; r(O2) = (1/2)r = 3.25×10^{-...}. Thus NO2 then O2 correspond to 1.3×... and 3.25×... respectively ⇒ (b).
Moles O2 formed per min = 48/32 = 1.5 mol min^{-1}. Reaction: 2 H2O2 → 2 H2O + O2. For each 1 mol O2, 2 mol H2O formed. So rate(H2O) = 2×1.5 = 3.0 mol min^{-1}. (d).
For zero order t1/2 = [A]0/(2k) ∝ [A]0. Doubling initial concentration doubles t1/2. So the reaction is zero order. (a).
For A → 3 gaseous products, if initial pressure P0 (only A) and at time t total pressure P, remaining A partial pressure pA = P0 e^{-kt}. Total pressure P = pA + products = P0 e^{-kt} + 3(P0 - P0 e^{-kt}) = P0(3 - 2e^{-kt}). Thus e^{-kt} = (3P0 - P)/(2P0). Hence k = -(1/t) ln[(3P0 - P)/(2P0)] = (2.303/t) log[2P0/(3P0 - P)].
For first order: fraction remaining after t: e^{-kt}. 75% completed → 25% remains: e^{-k·60} = 0.25 ⇒ k = ln4/60. Half-life t1/2 = ln2/k = ln2 /(ln4/60) = 60/2 = 30 min.
560 days = 4 half-lives (560/140 = 4). Remaining fraction = (1/2)^4 = 1/16. So 1 g → 1/16 g.
For first order t1/2 = ln2/k, independent of initial concentration. For second order (A → products) t1/2 = 1/(k[A]0), so it depends on [A]0. Options a, c, d are incorrect statements.
1/16 remaining corresponds to 4 half-lives. Total time 2 h = 120 min = 4·t1/2 ⇒ t1/2 = 120/4 = 30 min.
Average rate is the mean change over a finite time interval. Instantaneous rate is the limiting value as Δt → 0, given by the time derivative of concentration.
Average rate: change in concentration of a reactant or product divided by time interval Δt: average rate = −Δ[A]/Δt (for reactant) or +Δ[P]/Δt (for product). Instantaneous rate: rate at a specific time t; mathematically the derivative: instantaneous rate = −d[A]/dt (or d[P]/dt) evaluated at that time.
Rate law gives how rate depends on concentrations; rate constant sets the scale and includes Arrhenius dependence on temperature (k = A e^{-Ea/RT}).
Rate law: an expression relating the reaction rate to the concentrations of reactants (and sometimes products or catalysts), e.g. rate = k [A]^m [B]^n, where m,n are orders determined experimentally. Rate constant (k): proportionality constant in the rate law; depends on temperature and catalyst but not on reactant concentrations; units depend on overall order.
Integrated form linear: [A] vs t is straight line with slope −k.
For zero order: rate = −d[A]/dt = k (constant). Integrating: d[A] = −k dt ⇒ ∫_{[A]0}^{[A]} d[A] = −k ∫_{0}^{t} dt ⇒ [A] − [A]0 = −kt ⇒ [A] = [A]0 − kt. The half-life t1/2 satisfies [A]0/2 = [A]0 − k t1/2 ⇒ t1/2 = [A]0/(2k).
Thus for first order t1/2 = 0.693/k and does not depend on initial concentration.
Half-life (t1/2) is the time required for concentration of a reactant to fall to half its initial value. For first order: [A] = [A]0 e^{-kt}. Set [A] = [A]0/2 ⇒ 1/2 = e^{-k t1/2} ⇒ t1/2 = ln2 / k, independent of [A]0.
Table (concise): molecularity — theoretical, integer, per elementary step; order — experimental, may be fractional, sum of powers in rate law, applies to overall reaction.
Elementary reaction: a single-step reaction in which reactant molecules collide and directly form products; molecularity is the number of reactant species that collide in that elementary step (unimolecular, bimolecular, termolecular). Differences: molecularity is an integer and applies only to elementary steps; order is determined experimentally from the rate law, can be fractional, and is the sum of exponents of concentration terms in the rate law. Molecularity equals the order only for elementary reactions; order refers to the overall reaction rate dependence.
Identify slow step and write rate law from that step (or combine using pre-equilibrium if slow step involves intermediates).
Rate-determining step (RDS) is the slowest step in a reaction mechanism which controls the overall rate. Example: mechanism: (1) A + B → I (slow) (2) I + B → products (fast). Overall rate ≈ rate of step (1) → rate = k1 [A][B]. The RDS sets the observable rate law because faster steps equilibrate quickly relative to the slow step.
Use ln[A] = ln[A]0 − kt for linear representation.
For first order: [A] = [A]0 e^{-kt}. Plots: (i) ln[A] vs t is a straight line with slope −k and intercept ln[A]0. (ii) log10[A] vs t is straight with slope −k/2.303. (iii) [A] vs t is an exponential decay curve. From ln[A] vs t one obtains k from slope.
Orders are the exponents on concentrations in the rate law; overall order (a) = 3/2, (b) = 3.
(a) rate = k [x]^{3/2} (since zero order in y means [y]^0 = 1). (b) rate = k [NO]^2 [Br2]^1 = k [NO]^2 [Br2].
Energy diagram: lower peak (Ea) for catalysed path ⇒ larger k (Arrhenius), hence faster rate. Catalyst may be homogeneous or heterogeneous.
A catalyst increases reaction rate by providing an alternative pathway with lower activation energy (Ea) without being consumed. Example: decomposition of H2O2 is slow but is accelerated by MnO2; catalyst lowers Ea and increases fraction of collisions that lead to product.
Rate ∝ [A][B][L]. (i) [L]→4[L] ⇒ rate×4. (ii) [A],[B]→2[A],2[B] ⇒ rate×2×2 = ×4. (iii) [A]→(1/2)[A] ⇒ rate×1/2. (iv) [A]→(1/4)[A], [L]→4[L] ⇒ rate×(1/4)×4 = ×1 (no change).
(i) 4 times (ii) 4 times (iii) 1/2 times (iv) unchanged (1 times).
For dimerization 2A → A2, assuming rate of formation of dimer r = k [A]^2. Thus k = r/[A]^2 = (7.5×10^{-5})/(0.05)^2 = (7.5×10^{-5})/(2.5×10^{-3}) = 0.03 L mol^{-1} s^{-1}.
k = 0.03 L mol^{-1} s^{-1}
Rate law exponents add: 1 (x) + 1 (y) = 2. Since [z] is absent, its order is zero (no effect on rate).
Overall order = 2. Order with respect to z = 0 (z does not appear in rate law).
Collision theory explains temperature dependence qualitatively and leads to the Arrhenius form when combined with energetic considerations.
Collision theory: for a bimolecular reaction A + B → products, reaction rate ∝ collision frequency between A and B × fraction of collisions with energy ≥ Ea × fraction with correct orientation. Only collisions meeting energy and orientation requirements lead to reaction, so rate = Z_AB · f · e^{-Ea/RT}, where Z_AB is collision frequency and f is orientation factor.
The equation shows k increases with T and decreases with larger Ea. Two-point form: ln(k2/k1) = (Ea/R)(1/T1 − 1/T2).
Arrhenius equation: k = A e^{-Ea/(RT)}. Here k is the rate constant, A is the frequency (pre-exponential) factor related to collision frequency and orientation, Ea is the activation energy, R is the gas constant, and T is absolute temperature. In logarithmic form: ln k = ln A − Ea/(RT).
Since pressure halves in 1 min, t1/2 = 1 min = 60 s. For first order k = ln2 / t1/2 = 0.693 / 60 = 0.01155 s^{-1} ≈ 1.155×10^{-2} s^{-1}.
k = 1.155 × 10^{-2} s^{-1}
Zero order kinetics arise when the rate is independent of reactant concentration, commonly for heterogeneous catalytic reactions with saturated active sites. Typical textbook examples: 2NH3(g) → N2(g) + 3H2(g) on Pt (surface-saturated) and H2 + D2 exchange on a metal catalyst.
Examples: (i) Heterogeneous catalytic decomposition of ammonia on a platinum surface (when the surface is saturated). (ii) Surface-catalysed hydrogenation/exchange reactions on metals (e.g., H2 + D2 exchange on a metal surface) — under conditions of surface saturation these show zero order kinetics.
If true rate = k[A]^m[B]^n and [B] ≫ [A] so [B]≈[B]0 (constant), then rate = k[A]^m[B]0^n = k'[A]^m. For m=1 this becomes first order in A with k' = k[B]0^n. Example: acid-catalysed hydrolysis of ethyl acetate in large excess of water: rate ≈ k'[EtOAc].
A pseudo-first-order reaction is one that is higher order overall but one reactant is in large excess so its concentration is effectively constant; the rate law then reduces to first order in the limiting reactant. Example: hydrolysis of tert-butyl chloride in water, rate = k[tert‑BuCl][H2O]; with water in large excess, [H2O]≈constant, so rate = k' [tert‑BuCl] with k' = k[H2O], i.e. pseudo‑first order.
(i) Rusting is a surface-catalysed process often independent of bulk concentration → zero order. (ii) Radioactive decay follows dN/dt = −λN → first order. (iii) Rate law shows proportionality to [A] and [B] → orders 1 and 1, overall order = 2.
(i) Rusting of iron: typically zero order (heterogeneous surface process). (ii) Radioactive disintegration of 92U: first order. (iii) Given rate = k[A][B]: first order in A and first order in B, overall second order.
Use Arrhenius: k = A e^{-Ea/(RT)}. Convert Ea = 200 kJ mol^{-1} = 200000 J mol^{-1}. Exponent: −Ea/(RT) = −200000/(8.314×600) ≈ −40.106. e^{-40.106} ≈ 3.88×10^{-18}. Thus k = 1.6×10^{13} × 3.88×10^{-18} ≈ 6.21×10^{-5} s^{-1}.
k = 6.2 × 10^-5 s^-1 (approximately).
Compare expt 1 and 2: [x] doubles, [y] constant ⇒ rate doubles (0.15 → 0.30) ⇒ first order in x. Compare expt 2 and 3: [y] increases 4× (0.02→0.08), [x] constant ⇒ rate increases 4× (0.30→1.20) ⇒ first order in y. Hence rate = k[x][y]. From expt 1: 0.15 = k(0.2)(0.02) = k(0.004) ⇒ k = 0.15/0.004 = 37.5 M^{-1}s^{-1}.
Rate law: rate = k [x]^1 [y]^1 = k[x][y], with k = 37.5 M^-1 s^-1.
For a rate law rate = k[A]^m[B]^n, the instantaneous rate changes as powers m,n of concentrations. Higher concentration → more frequent effective collisions → higher rate; reaction order (m,n) quantifies sensitivity of rate to concentration changes.
Rate depends on reactant concentrations according to the rate law: rate ∝ ∏[Reactant]^order. Increasing concentration increases collision frequency and hence usually increases rate; the exact dependence (order) must be determined experimentally.
Examples: C–C single bond cleavage is slower than reactions involving polar bonds; reactions between ions in solution are usually faster than between neutral molecules; solids react slower unless finely divided (larger surface area). Molecular orientation and steric hindrance also influence rate.
Nature affects rate via bond strengths, molecular structure, phase (gas/liquid/solid), and electronic factors: weaker bonds and more reactive functional groups react faster. Physical state and surface area (for solids) and presence of ionic vs covalent bonds also matter.
For first order t1/2 = ln2 / k = 0.693 / (1.54×10^{-3} s^{-1}) = 450 s.
t1/2 = 450 s (approximately).
k = ln2 / t1/2 = 0.693 / 8.0 min = 0.086625 min^{-1}. For first order, [A] = [A]0 e^{-kt}; require [A]/[A]0 = 0.01 ⇒ t = −ln(0.01)/k = 4.60517 / 0.086625 ≈ 53.2 min.
Approximately 53.2 minutes.
k = ln2 / t1/2 = 0.693 / 60 s = 0.01155 s^{-1}. After 180 s (three half-lives), fraction remaining = (1/2)^3 = 1/8 = 0.125 = 12.5%. Using exponential form: [A] = [A]0 e^{-kt} = [A]0 e^{-0.01155×180} ≈ 0.125 [A]0.
k = 0.01155 s^-1. After 180 s, 12.5% of A remains (i.e. 1/8 of initial).
Zero order: [A]t = [A]0 − kt. 20% complete ⇒ 0.2[A]0 = kt (at t = 20 min) ⇒ k = 0.2[A]0 / 20 = 0.01[A]0 min^{-1}. For 80% complete need 0.8[A]0 = k t ⇒ t = 0.8[A]0 / k = 0.8[A]0 / (0.01[A]0) = 80 min.
k = (0.2[A]0)/20 min = 0.01[A]0 min^-1. Time for 80% completion = 80 min.
Arrhenius: k = A e^{-Ea/(RT)} ⇒ A = k e^{Ea/(RT)}. Convert Ea = 22.5 kcal mol^{-1} = 22.5×4184 = 94140 J mol^{-1}. T = 40°C = 313 K. Ea/(RT) = 94140/(8.314×313) ≈ 36.16. e^{36.16} ≈ 5.075×10^{15}. Thus A = 1.8×10^{-5} × 5.075×10^{15} ≈ 9.1×10^{10} s^{-1}.
A ≈ 9.1 × 10^10 s^-1.
For a first order gas-evolving reaction, V∞ − Vt ∝ [A]t. Take V∞ = 58.3 ml. Compute ln(V∞ − Vt): For volumes 19.3,32.6,41.3,46.5,50.4 → differences 39.0,25.7,17.0,11.8,7.9 with ln values ≈3.663,3.246,2.833,2.468,2.067. Plot ln(V∞−Vt) vs t (corresponding times typically 0,5,10,15,20 min) — points lie on a straight line. Slope = −k; using e.g. (3.663−2.067)/(0−20) = −0.0798 min^{-1}. Other intervals give ~0.078–0.083 min^{-1}. Average k ≈ 0.080 min^{-1}. Thus reaction is first order with k ≈ 8.0×10^{-2} min^{-1}.
The data fit first order: ln(V∞ − Vt) vs t is linear. Estimated k ≈ 8.0 × 10^-2 min^-1.
For first order, ln V vs t should be linear (V ∝ concentration). Compute ln V: ln(46.1)=3.830, ln(29.8)=3.396, ln(19.3)=2.960. Differences per 10 min: (3.830−3.396)/10 = 0.0434 and (3.396−2.960)/10 = 0.0436 min^{-1}. These equal k ≈ 0.0435 min^{-1}. Thus first order with k ≈ 4.35×10^{-2} min^{-1}.
The reaction is first order. k ≈ 4.35 × 10^-2 min^-1.
40% complete ⇒ fraction remaining = 0.60. For first order k = −(1/t) ln([A]t/[A]0) = −(1/50) ln(0.60) = (0.5108)/50 = 0.010216 min^{-1}. For 80% complete, remaining fraction = 0.20, so t = −ln(0.20)/k = 1.6094 / 0.010216 ≈ 157.6 min ≈ 158 min.
k ≈ 1.0216 × 10^-2 min^-1. Time for 80% completion ≈ 158 minutes.