Exercises — Methods to determine order — initial rates, integrated method, half-life method

WHAT: find $n$ from $r_2/r_1=([A{]}_2/[A{]}_1{)}^n$. WHY log: the $n$ is stuck as an exponent; $\ln$ brings it down. Ratio of rates $=8$, ratio of concentrations $=2$, so $2^n=8$. Since $8=2^3$, we read off $n=3$. Or formally: $n=\frac{\ln 8}{\ln 2}=3$. Answer: third order.

Only the first-order integrated law, $\ln [A]=\ln [A{]}_0-kt$, is linear when you plot $\ln [A]$ against $t$ (it has the shape "constant $-$ constant $\times t$"). Answer: first order.

Recall the fingerprint $t_{1/2}\propto [A{]}_0^{1-n}$. "Unchanged" means the exponent $1-n=0$, so $n=1$. Answer: first order. (This is the radioactivity case — see Radioactive Decay.)

Step 1 — rate ratio: $\frac{1.08\times 10^{-2}}{1.2\times 10^{-3}}=9$. Step 2 — concentration ratio: $\frac{0.150}{0.050}=3$. Step 3 — solve the exponent: $3^n=9\Rightarrow n=\frac{\ln 9}{\ln 3}=2$. Second order. Step 4 — get $k$ from $r=k[A{]}^2$ using Expt 1: $k=\frac{r}{[A{]}^2}=\frac{1.2\times 10^{-3}}{(0.050{)}^2}=0.48{\text{M}}^{-1}{\text{s}}^{-1}.$

WHY first-order law: given as first order, so $\ln [A]=\ln [A{]}_0-kt$ (see Integrated Rate Equations). Find $k$: $k=\frac{1}{t}\ln \frac{[A{]}_0}{[A]}=\frac{1}{60}\ln \frac{0.80}{0.20}=\frac{\ln 4}{60}=0.02310{\text{s}}^{-1}.$ At $t=120$ s: notice $120$ s is two 60-s intervals, and each 60 s dropped $[A]$ to a quarter (0.80→0.20). So after another 60 s: $0.20\to 0.050$ M. Check with the law: $[A]=0.80e^{-0.02310\times 120}=0.80\times e^{-2.772}=0.050$ M. ✓ Answers: $k=0.0231{\text{s}}^{-1}$, $[A{]}_{120}=0.050$ M.

Find $a$ — compare 1→2 (here $[B]$ is held fixed, so only $A$'s power can move): rate ratio $=4.0/2.0=2$; concentration ratio $=0.20/0.10=2$; so $2^a=2\Rightarrow a=1$. Find $b$ — compare 2→3 ($[A]$ fixed at 0.20, only $B$ changes): rate ratio $=3.6\times 10^{-3}/4.0\times 10^{-4}=9$; concentration ratio $=0.30/0.10=3$; so $3^b=9\Rightarrow b=2$. Overall order $=a+b=3$. Find $k$ using Expt 1: $r=k[A][B{]}^2$ $k=\frac{2.0\times 10^{-4}}{(0.10)(0.10{)}^2}=\frac{2.0\times 10^{-4}}{1.0\times 10^{-3}}=0.20{\text{M}}^{-2}{\text{s}}^{-1}.$

Look at the third column $1/[A]$: values $1,2,3,4$ rise by a constant $+1$ every $10$ s — a perfectly straight line. The $\ln [A]$ column jumps by $-0.693,-0.406,-0.287$ (shrinking), so not linear. Straight $1/[A]$ vs $t$ $\Rightarrow$ second order. Slope $=\frac{4.00-1.00}{30-0}=0.10{\text{M}}^{-1}{\text{s}}^{-1}$, and for $n=2$ the slope is $+k$, so $k=0.10{\text{M}}^{-1}{\text{s}}^{-1}$. See figure below.

Step 1 — order from the fingerprint $t_{1/2}\propto [A{]}_0^{1-n}$: $n=1-\frac{\ln (t_{1/2,2}/t_{1/2,1})}{\ln ([A{]}_{0,2}/[A{]}_{0,1})}=1-\frac{\ln (10/40)}{\ln (0.80/0.20)}=1-\frac{\ln (1/4)}{\ln 4}=1-(-1)=2.$ Second order. (Sanity: $[A{]}_0$ went ×4, $t_{1/2}$ went ÷4 — inverse, exactly the $n=2$ signature.) Step 2 — get $k$ from the second-order half-life $t_{1/2}=\frac{1}{k[A{]}_0}$, so $k=\frac{1}{t_{1/2}[A{]}_0}$. Using the first row: $k=\frac{1}{40\times 0.20}=0.125{\text{M}}^{-1}{\text{s}}^{-1}$. Check with second row: $\frac{1}{10\times 0.80}=0.125$. ✓ Consistent.

Order from initial rates: rate ratio $=8.0\times 10^{-2}/5.0\times 10^{-3}=16$; concentration ratio $=0.40/0.10=4$; so $4^n=16\Rightarrow n=2$. Second order. $k$ from rate law: $k=\frac{r}{[A{]}^2}=\frac{5.0\times 10^{-3}}{(0.10{)}^2}=0.50{\text{M}}^{-1}{\text{s}}^{-1}$. $k$ from half-life (second order, $t_{1/2}=1/k[A{]}_0$): $k=\frac{1}{t_{1/2}[A{]}_0}=\frac{1}{200\times 0.10}=0.050{\text{M}}^{-1}{\text{s}}^{-1}$. Wait — mismatch! $0.50\ne 0.050$. The two methods disagree, which by the parent's rule means a data/unit error, not two different "true" orders. Re-checking: if instead $t_{1/2}=20$ s, then $k=1/(20\times 0.10)=0.50$ — matching. So the consistent dataset is $t_{1/2}=20$ s, and both methods give $k=0.50{\text{M}}^{-1}{\text{s}}^{-1}$, second order. Lesson: all methods must return the same $k$; a clash flags an error.

Step 1 — spot the pattern: $[A]$ drops by a constant $0.25$ M every $50$ s. A constant amount per unit time is the signature of $[A]=[A{]}_0-kt$ → zero order (linear $[A]$ vs $t$). Neither $\ln [A]$ nor $1/[A]$ would be straight here. Step 2 — $k$: slope of $[A]$ vs $t$ is $\frac{0.25-1.00}{150-0}=-5.0\times 10^{-3}$ M/s. For $n=0$ slope $=-k$, so $k=5.0\times 10^{-3}$ M/s. Step 3 — $t_{1/2}$ at $[A{]}_0=1.00$: zero-order $t_{1/2}=\frac{[A{]}_0}{2k}=\frac{1.00}{2(5.0\times 10^{-3})}=100$ s. (Matches: $[A]$ hit $0.50$ at $t=100$ s. ✓) Step 4 — restart at $0.40$ M: $t_{1/2}=\frac{0.40}{2(5.0\times 10^{-3})}=40$ s. Smaller start → shorter half-life, because zero-order $t_{1/2}\propto [A{]}_0$. See the plot below.

$t_{1/2}$ at 300 K (first order): $t_{1/2}=\frac{\ln 2}{k}=\frac{0.693}{0.010}=69.3$ s — and note it does not depend on $[A{]}_0$. After $k$ doubles: $t_{1/2}^{\prime }=\frac{0.693}{0.020}=34.65$ s. Factor: $t_{1/2}^{\prime }/t_{1/2}=\frac{1}{2}$ — the half-life is halved, because for first order $t_{1/2}\propto 1/k$. Why $k$ rose with temperature: that is the Arrhenius Equation at work ($k=A{e}^{-E_a/RT}$); a faster $k$ always means a shorter half-life here.

Initial rates → $n=\frac{\ln (r_2/r_1)}{\ln ([A{]}_2/[A{]}_1)}$; integrated → which of $[A],\ln [A],1/[A]$ is straight; half-life → sign of $1-n$ in $t_{1/2}\propto [A{]}_0^{1-n}$. All three must agree on one $n$ and one $k$.