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 k = (2.303/t) log([R]0/[R]) with log 5 - log 4 = 0.1 to get k, then t = 70 min. A straight ln[R] vs t line means first order with slope -k.

Answer 1

Step 1: Write the first order rate law:
For a first order reaction, the integrated equation is:
\[ k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \]
Take the initial amount \( [R]_0 = 100 \).

Step 2: Find k from the 20% data:
After 10 min, 20% has reacted, so \( [R] = 80 \).
\[ k = \frac{2.303}{10}\log\frac{100}{80} = \frac{2.303}{10}\log\frac{5}{4} \]
\[ \log\frac{5}{4} = \log 5 - \log 4 = 0.7 - 0.6 = 0.1 \]
\[ k = 0.2303 \times 0.1 = 0.02303\ \mathrm{min^{-1}} \]

Step 3: Find the time for 80% completion:
When 80% has reacted, \( [R] = 20 \).
\[ t = \frac{2.303}{0.02303}\log\frac{100}{20} = 100 \times \log 5 = 100 \times 0.7 = 70\ \mathrm{min} \]

Step 4: Order from the graph (part ii, I):
The graph of \( \ln[R] \) against t is a straight line sloping down.
This matches \( \ln[R] = \ln[R]_0 - kt \), which is the first order equation. So the reaction is first order.

Step 5: Slope of the line (part ii, II):
Compare \( \ln[R] = -kt + \ln[R]_0 \) with \( y = mx + c \). The slope is \( m = -k \).

Final Answer:
(i) Time for 80% completion = 70 minutes.
(ii) (I) First order. (II) Slope = \( -k \).