Question

(i) A first order reaction is 20% complete in 10 minutes. Calculate the time taken when 80% of the reaction will be completed.

[Given : log 5 = 0.7, log 4 = 0.6]

(ii) For a chemical reaction \( \mathrm{R \rightarrow P} \), variation in \( \ln[\mathrm{R}] \) vs. time (t) graph is given below :

(I) Predict the order of the reaction.

(II) Write the slope of the curve.

Hint:

Use the first order law for each stage and note that the ln[R] vs t line has slope -k.

Answer 1

Step 1: Part (i) - Find the rate constant:
For a first order reaction the rate constant is given by the integrated rate law.
\[ k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \]
When the reaction is 20% complete, 80% of the reactant is left. So \( [R]_0 = 100 \) and \( [R] = 80 \).
\[ k = \frac{2.303}{10\ \mathrm{min}}\log\frac{100}{80} = \frac{2.303}{10}\log\frac{5}{4} \]
Now \( \log\frac{5}{4} = \log 5 - \log 4 = 0.7 - 0.6 = 0.1 \).
\[ k = \frac{2.303 \times 0.1}{10} = 0.02303\ \mathrm{min^{-1}} \]

Step 2: Part (i) - Time for 80% completion:
When 80% is over, 20% is left. So \( [R]_0 = 100 \) and \( [R] = 20 \).
\[ t = \frac{2.303}{k}\log\frac{100}{20} = \frac{2.303}{0.02303}\log 5 \]
\[ t = 100 \times 0.7 = 70\ \mathrm{min} \]

Step 3: Part (ii)(I) - Order from the graph:
The integrated rate law of a first order reaction is \( \ln[R] = \ln[R]_0 - kt \).
This has the form of a straight line, \( y = c + mx \), with \( \ln[R] \) on the y-axis and t on the x-axis.
The given graph is a straight line that starts at \( \ln[R]_0 \) and falls with time. So the reaction is of first order.

Step 4: Part (ii)(II) - Slope of the line:
Comparing the two equations, the slope \( m = -k \).
The line falls, so the slope is negative. Its magnitude is the rate constant.

Final Answer:
(i) The time for 80% completion is 70 minutes. (ii) The reaction is of first order and the slope is \( -k \).
\[ \boxed{t_{80\%} = 70\ \mathrm{min};\ \text{first order};\ \text{slope} = -k} \]