Q.Derive integrated rate law for a zero order reaction $A \rightarrow \text{product}$.

Step 1. For a zero order reaction $A\rightarrow\text{product}$, $\text{Rate}=k[A]^0=k$, so $-\dfrac{d[A]}{dt}=k\Rightarrow-d[A]=k\,dt$.

Step 2. Integrate between $t=0$ (where $[A]=[A]_0$) and $t=t$ (where $[A]=[A]$): $-\displaystyle\int_{[A]_0}^{[A]}d[A]=k\int_0^t dt$.

Step 3. Evaluating: $-\big([A]-[A]_0\big)=k(t-0)\Rightarrow[A]_0-[A]=kt$.

Step 4. Solving for k: $k=\dfrac{[A]_0-[A]}{t}$. Rearranged as $[A]=-kt+[A]_0$, this matches $y=mx+c$, so a plot of $[A]$ against t is a straight line of slope $-k$ and intercept $[A]_0$.

Set up the differential rate law for zero order, integrate both sides between the initial and later states, and identify the resulting straight-line form.

• Attempting to integrate $\int d[A]/[A]$ (the FIRST order procedure) instead of the simpler $\int d[A]$ appropriate for zero order, since $[A]^0=1$ has no concentration term to divide through by.