Chemical Kinetics: questions with answers

The reaction consumes 3 mol of hydrogen for every 2 mol of ammonia formed. Therefore, $\frac{d[\mathrm{NH_3}]}{dt}=\frac{2}{3}\left(-\frac{d[\mathrm{H_2}]}{dt}\right)=\frac{2}{3}\times0.060=0.040\ \mathrm{mol\,L^{-1}\,s^{-1}}$.

The stoichiometric rate expression is $-\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt} = \frac{1}{4}\frac{d[\mathrm{NO_2}]}{dt}$. Thus, the disappearance rate of $\mathrm{N_2O_5}$ is $\frac{2}{4} \times 4.0 \times 10^{-4} = 2.0 \times 10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}}$.

The stoichiometric coefficient of $H_2O_2$ is 2. Divide its disappearance rate by 2 to obtain $r=-\frac{1}{2}\frac{d[H_2O_2]}{dt}$. The negative sign makes the reaction rate positive as the reactant concentration decreases.

Overall order is the sum of the concentration exponents in the experimental rate law. In the second medium, the order is $1/2+1=3/2$. The order of $2$ applies to the first medium.

Doubling the volume halves both reactant concentrations. The rate constant remains unchanged at constant temperature. Therefore, $r_{\mathrm{new}}/r_{\mathrm{initial}}=(1/2)^2(1/2)=1/8$.

Water is present in large excess, so its concentration barely changes. With the acid catalyst concentration fixed, the rate law becomes $\text{rate}=k'[sugar]$. The reaction is therefore pseudo first order, with observed order 1.

The Arrhenius equation gives $\ln k=\ln A-E_a/(RT)$, so the slope is $-E_a/R$. Halving the activation energy halves the slope's magnitude. The new slope is $(-4000)/2=-2000\ \mathrm{K}$.

A catalyst does not change the equilibrium constant at constant temperature. Since $K = k_f/k_b$ remains unchanged, increasing $k_f$ by a factor of 100 requires $k_b$ to increase by the same factor.

For ln k against 1/T, slope = −Eₐ/R. Thus, Eₐ = −R × slope = 8.314 × 5000 = 41570 J mol⁻¹ ≈ 41.6 kJ mol⁻¹.

For a zero-order reaction, $t_{1/2}=[A]_0/(2k)$. At unchanged $k$, doubling the initial concentration doubles the half-life. The new half-life is $60\times(0.20/0.10)=120\ \mathrm{s}$.

For a first-order reaction, $t_{1/2}=0.693/k$. Doubling the rate constant halves the half-life. The new half-life is $60/2=30\ \text{s}$.

The first order integrated rate equation is $\ln[R] = \ln[R]_0 - kt$. Thus, ln[R] versus t gives a straight line with slope $-k$ and intercept $\ln[R]_0$.

With collision frequency and the Boltzmann energy factor constant, the reaction rate is proportional to the steric factor. Thus, $r_{\text{new}}/r_{\text{original}}=0.25/0.50=1/2$.

Reaction requires collisions with sufficient energy and suitable orientation. Collision frequency alone is not the rate: the Boltzmann energy factor and steric probability determine the fraction of effective collisions.

The energy fraction is $e^{-E_a/(RT)}$. Lowering $E_a$ at fixed $T$ decreases $E_a/(RT)$. The exponent becomes less negative, so the fraction increases.