Solved Examples (Dedicated Problem Set)

How to Use This Problem Set

This is your full workout for Chemical Kinetics, grouped by theme: rate and rate relations, order and units of $k$, zero- and first-order integrated-rate calculations, half-life, pseudo-first-order, and Arrhenius / activation-energy problems.

Solve each with a pen before reading the solution. Keep these handy:

• First order: $k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$; $t_{1/2} = \dfrac{0.693}{k}$.
• Zero order: $[R] = [R]_0 - kt$; $t_{1/2} = \dfrac{[R]_0}{2k}$.
• Arrhenius: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{T_2-T_1}{T_1 T_2}\right)$, $R = 8.314$ J K$^{-1}$ mol$^{-1}$.

Solved Examples — Rate & Order

Example 1. For $2\text{A} \rightarrow$ products, $[A]$ falls 0.50 to 0.40 M in 10 min. Find the rate of reaction.

Solution: Rate $= -\tfrac{1}{2}\dfrac{\Delta[A]}{\Delta t} = \tfrac{1}{2}\times\dfrac{0.10}{10} = 5\times10^{-3}$ mol L$^{-1}$ min$^{-1}$.

Example 2. Rate $= k[A]^2[B]$. Overall order?

Solution: $2 + 1 = 3$ (third order).

Example 3. A rate constant has units L mol$^{-1}$ s$^{-1}$. Order?

Solution: These are the units of $k$ for second order.

Solved Examples — Order & Units

Example 4. For Rate $= k[A]^{1/2}[B]^{3/2}$, find the overall order.

Solution: $\tfrac{1}{2} + \tfrac{3}{2} = 2$ (second order).

Example 5. Doubling $[A]$ in Rate $= k[A]^2$ changes the rate by what factor?

Solution: $2^2 = 4$ times.

Example 6. For Rate $= k[A][B]^2$, doubling both concentrations changes the rate by:

Solution: $2^1\times2^2 = 8$ times.

Solved Examples — Zero Order

Example 7. Zero order, $[R]$ falls 0.10 to 0.075 M in 50 s. Find $k$.

Solution: $k = \dfrac{0.10-0.075}{50} = 5\times10^{-4}$ mol L$^{-1}$ s$^{-1}$.

Example 8. Zero order, $k = 0.02$ mol L$^{-1}$ s$^{-1}$, $[R]_0 = 0.10$ M. Find $t_{1/2}$.

Solution: $t_{1/2} = \dfrac{[R]_0}{2k} = \dfrac{0.10}{0.04} = 2.5$ s.

Example 9. Zero order, $k = 0.05$ mol L$^{-1}$ s$^{-1}$, $[R]_0 = 0.80$ M. Time to reach 0.50 M?

Solution: $t = \dfrac{0.80-0.50}{0.05} = 6$ s.

Solved Examples — First Order

Example 10. First order, 50% complete in 20 min. Find $k$.

Solution: $k = \dfrac{2.303}{20}\log2 = 0.0347$ min$^{-1}$.

Example 11. First order, $k = 2.303\times10^{-3}$ s$^{-1}$. Fraction left after 1000 s?

Solution: $\log\dfrac{[R]_0}{[R]} = \dfrac{kt}{2.303} = 1$, so ratio = 10; fraction = 0.1 (10%).

Example 12. First order, $[R]$ falls 0.80 to 0.20 M in 60 s. Find $k$.

Solution: $k = \dfrac{2.303}{60}\log4 = 0.0231$ s$^{-1}$.

Solved Examples — First Order (continued)

Example 13. First order takes 40 min for 30% decomposition. Find $k$.

Solution: $k = \dfrac{2.303}{40}\log\dfrac{100}{70} = 8.92\times10^{-3}$ min$^{-1}$.

Example 14. Show $t_{99\%} = 2\,t_{90\%}$ for first order.

Solution: $t_{99} = \dfrac{2.303}{k}\log100 = \dfrac{2.303}{k}(2)$; $t_{90} = \dfrac{2.303}{k}(1)$. Ratio = 2.

Example 15. First order, $k = 1.15\times10^{-3}$ s$^{-1}$. Time to fall to one-fourth of $[R]_0$?

Solution: $t = \dfrac{2.303}{1.15\times10^{-3}}\log4 = 1206$ s.

Solved Examples — Half-Life

Example 16. First order, $k = 1.386\times10^{-3}$ s$^{-1}$. Find $t_{1/2}$.

Solution: $t_{1/2} = \dfrac{0.693}{1.386\times10^{-3}} = 500$ s.

Example 17. $^{14}$C ($t_{1/2} = 5730$ yr); a sample has 25% left. Age?

Solution: 25% = 2 half-lives = $2\times5730 = 11460$ yr.

Example 18. First order is 75% complete in 60 min. Find $t_{1/2}$.

Solution: 75% complete means 25% remains, which is 2 half-lives in 60 min. Therefore, $t_{1/2} = 30$ min.

Solved Examples — Half-Life (continued)

Example 19. First order, $t_{1/2} = 10$ min. Fraction left after 30 min?

Solution: $30/10 = 3$ half-lives, fraction $= (1/2)^3 = 1/8$ (12.5%).

Example 20. First order, $t_{1/2} = 69.3$ s. Find $k$.

Solution: $k = \dfrac{0.693}{69.3} = 0.01$ s$^{-1}$.

Example 21. $^{90}$Sr ($t_{1/2} = 28.1$ yr). Fraction left after 84.3 yr?

Solution: $84.3/28.1 = 3$ half-lives, fraction = $1/8$ (12.5%).

Solved Examples — Arrhenius

Example 22. Rate doubles 298 -> 308 K. Find $E_a$ ($R = 8.314$).

Solution: $\log2 = \dfrac{E_a}{2.303R}\dfrac{10}{298\times308} \Rightarrow E_a = 52.9$ kJ mol$^{-1}$.

Example 23. Arrhenius plot slope = $-5000$ K. Find $E_a$.

Solution: $E_a = 5000\times8.314 = 41570$ J $= 41.6$ kJ mol$^{-1}$.

Example 24. $k$ rises $1.0\times10^{-3}$ to $4.0\times10^{-3}$ s$^{-1}$, 300 -> 320 K. Find $E_a$.

Solution: $\log4 = \dfrac{E_a}{2.303R}\dfrac{20}{300\times320} \Rightarrow E_a = 55.3$ kJ mol$^{-1}$.

Solved Examples — Arrhenius (continued)

Example 25. $k_1 = 2.0\times10^{-3}$ s$^{-1}$ at 300 K, $E_a = 50$ kJ mol$^{-1}$. Find $k_2$ at 310 K.

Solution: $\log\dfrac{k_2}{k_1} = \dfrac{50000}{2.303\times8.314}\dfrac{10}{300\times310} = 0.281$; ratio = 1.91; $k_2 = 3.82\times10^{-3}$ s$^{-1}$.

Example 26. $E_a = 0$ in the Arrhenius equation implies?

Solution: $k = A$, independent of temperature.

Example 27. A catalyst lowers $E_a$ from 75 to 50 kJ mol$^{-1}$. Effect on $k$?

Solution: $k$ increases (smaller $E_a$ in $k = A e^{-E_a/RT}$).

Solved Examples — Mixed

Example 28. For $2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2$, $d[\text{O}_2]/dt = 1.0\times10^{-3}$ mol L$^{-1}$ s$^{-1}$. Find $d[\text{NO}_2]/dt$.

Solution: $d[\text{NO}_2]/dt = 4\times d[\text{O}_2]/dt = 4.0\times10^{-3}$ mol L$^{-1}$ s$^{-1}$.

Example 29. A reaction is second order in A. If $[A]$ is halved, the rate becomes:

Solution: Rate $\propto [A]^2$, so $(1/2)^2 = 1/4$ of the original.

Example 30. First order, $k = 0.0693$ min$^{-1}$, $[R]_0 = 1.0$ M. $[R]$ after 10 min?

Solution: $\log\dfrac{[R]_0}{[R]} = \dfrac{0.0693\times10}{2.303} = 0.301$; ratio = 2; $[R] = 0.50$ M.

Example 31. Decomposition of NH$_3$ on hot Pt is zero order. Why?

Solution: The Pt surface is saturated, so rate is independent of $[\text{NH}_3]$.

Example 32. Units of $k$ for a third-order reaction?

Solution: mol$^{-2}$ L$^2$ s$^{-1}$.