Explanation:

For this reaction, r = −(1/2)Δ[N₂O₅]/Δt = (1/4)Δ[NO₂]/Δt = Δ[O₂]/Δt. The reactant concentration decreases, while product concentrations increase. Only E uses the correct sign and stoichiometric factor.

Explanation:

The stoichiometric relation is r = −d[W]/dt = −(1/2)d[X]/dt = (1/3)d[Y]/dt = d[Z]/dt. Thus Y forms three times as fast as Z, and all concentration-change rates have the same units, mol/(L·s). Stoichiometry alone does not specify how the reaction rate changes with time.

Explanation:

Common factors affecting reaction rates include the chemical nature and concentration of reactants, temperature, catalysts, and the exposed surface area of solid reactants. Under the stated assumptions, product concentration does not affect the forward rate. In more complex or reversible reactions, products can influence the observed rate.

Explanation:

Warming increases the fraction of sufficiently energetic molecules and, for a positive activation energy, increases the Arrhenius rate constant. The activation-energy barrier itself need not increase. A doubling for every 10 °C rise is only a rough rule for some reactions, not a universal law.

Explanation:

At a fixed temperature and preexponential factor, a higher activation energy gives a smaller rate constant according to k = Ae^−Eₐ/(RT). Covalent bonding by itself does not establish whether a reaction is slow; the reaction pathway and its energy barrier matter.

Explanation:

The rate ratio is (2)²(2) = 8. Therefore, the rate increases by a factor of 8, and E is correct.

Explanation:

Write rate = k[A]^m[B]ⁿ. Comparing trials 2 and 3, B is constant and doubling A doubles the rate, so m = 1. Comparing trials 1 and 2, A is constant and tripling B increases the rate ninefold, so n = 2. Thus rate = k[A][B]² (A).

Explanation:

Comparing trials 1 and 2, [B] is constant while [A] doubles and the rate quadruples, so m = 2. From trials 2 to 3, both concentrations double but the rate increases only fourfold: 2² × 2ⁿ = 4. Thus n = 0, rate = k[A]², and the overall order is 2 + 0 = 2 (A). All three trials give k = 1600 M⁻¹·s⁻¹.

Explanation:

The overall order is 3. Therefore, [k] = (M·s⁻¹)/M³ = M⁻²·s⁻¹, which is D.

Explanation:

From trials 1 and 2, doubling Y at constant X quadruples the rate, so the order in Y is 2. From trials 2 and 3, tripling X at constant Y increases the rate ninefold, so the order in X is 2. Therefore rate = k[X]²[Y]². Using trial 1, k = 1/(10² × 20²) = 2.5 × 10⁻⁵ M⁻³·s⁻¹ (E). Each trial gives the same k.

Explanation:

The mass decreases through three half-lives: 40.0 → 20.0 → 10.0 → 5.00 g. The elapsed time is 3 × 7.0 min = 21 min (A).

Explanation:

The elapsed time is 200/50 = 4 half-lives. The remaining fraction is (1/2)⁴ = 1/16 = 0.0625, or 6.25% (B).

Explanation:

For a first-order reaction, t_1/2 = ln 2/k. Thus k = 0.693/(3.0 min) = 0.231 min⁻¹ ≈ 0.23 min⁻¹ (B).

Explanation:

The integrated first-order rate law is ln[A] = ln[A]₀ − kt. A plot of ln[A] against time has a negative slope, −k, so D is incorrect. Its half-life is t_1/2 = ln 2/k.

Explanation:

For a second-order reaction in A, 1/[A] = 1/[A]₀ + kt. Thus 1/[A] versus time is a straight line with positive slope, as in C. A and E are the characteristic linear plots for zero- and first-order reactions, respectively. A decreasing, curved ln[A] plot such as D can be qualitatively compatible with second-order decay, but it is not the requested linear plot.

Explanation:

The transition-state energy is 100 + 50 = 150 kJ. Subtract the reverse activation energy to obtain the product energy: 150 − 90 = 60 kJ. The reaction-energy change is 60 − 100 = −40 kJ; this is distinct from the product energy.

Explanation:

A catalyst provides a faster reaction pathway and accelerates both directions. It does not change the reaction’s standard Gibbs free-energy change or the equilibrium constant at a fixed temperature, so E is incorrect. It need not increase the total number of collisions.

Explanation:

The reaction rate depends on both k and [A]ᵐ. With the stated positive activation energy, warming increases k; with m > 0, increasing [A] increases the rate. At a fixed temperature and unchanged mechanism and catalyst conditions, the ideal-model rate constant is independent of [A]. All four statements are correct, so choose E.

Explanation:

Add the three steps and cancel B, 2BD, and F, which occur on both sides. The fully simplified net reaction is 2A + 2C → 3E (A). E is an unsimplified sum containing canceling species, not the requested final net equation.

Explanation:

Adding the steps gives A + D → BD + E. B is consumed in step 1 and regenerated in step 2, so it acts as a catalyst. C is formed and then consumed, so it is an intermediate. For elementary step 1, rate = k₁[A][B]. Under the explicitly equal-prefactor assumption at the same temperature, k₁ < k₂ implies Eₐ,₁ > Eₐ,₂, so E is incorrect. Slow/fast labels alone would not establish this energy ordering.