Step 1: Understanding the Concept: For a first order reaction the rate constant is found from the integrated rate law. Once k is known, it can be used to find the time for any other extent of reaction.
Step 2: Key Formula or Approach: \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \] If 20% is complete, 80% of the reactant is left. If 80% is complete, only 20% is left.
Step 3: Find the rate constant (part i): Let the initial amount be 100. After 10 minutes, the amount left is 100 - 20 = 80. \[ k = \frac{2.303}{10} \log \frac{100}{80} = \frac{2.303}{10} \log \frac{5}{4} \] Now \( \log \frac{5}{4} = \log 5 - \log 4 = 0.7 - 0.6 = 0.1 \). \[ k = \frac{2.303}{10} \times 0.1 = 0.02303 \ \mathrm{min^{-1}} \]
Step 4: Find the time for 80% completion (part i): When 80% is complete, the amount left is 20. \[ t = \frac{2.303}{k} \log \frac{100}{20} = \frac{2.303}{0.02303} \log 5 \] \[ t = 100 \times 0.7 = 70 \ \mathrm{min} \]
Step 5: Order from the graph (part ii, I): The integrated first order law can be written as \( \ln[R] = \ln[R]_0 - kt \). This has the form y = c + mx, so a plot of \( \ln[R] \) against t is a straight line only for a first order reaction. The graph shown is a straight line falling from \( \ln[R]_0 \), so the reaction is of first order.
Step 6: Slope of the curve (part ii, II): Comparing \( \ln[R] = \ln[R]_0 - kt \) with y = mx + c, the slope m is the coefficient of t. So the slope is \( -k \). It is negative because \( [R] \) falls with time.
Final Answer: (i) The time for 80% completion is 70 minutes. (ii) The order is one, and the slope is \( -k \). \[ \boxed{t = 70\ \text{min};\ \text{order} = 1;\ \text{slope} = -k} \]
