Consider a first order reaction
Match List-I with List-II
| List-I | List-II |
| Plot of | Type of graph |
| (A) [R] vs t | (I) Straight line passing through origin with positive slope (k/2.303) |
| (B) ln [R] vs t | (II) Exponential decrease |
| (C) ln ([R]0/[R]) vs t | (III) Straight line with negative slope (-k) |
| (D) log ([R]0/[R]) vs t | (IV) Straight line passing through origin with positive slope (k) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
All four plots come from the integrated rate equation of a first order reaction, written in different forms:
\( \displaystyle k=\frac{1}{t}\ln\frac{[R]_0}{[R]} \)
(A) [R] vs t — This matches (II) Exponential decrease. The concentration falls exponentially with time, quickly at first and then more slowly.
\( \displaystyle [R]=[R]_0\,e^{-kt} \)
(B) ln [R] vs t — This matches (III) Straight line with negative slope (-k). The equation has the form y = c + mx with intercept ln [R]0 and slope −k.
\( \displaystyle \ln[R]=\ln[R]_0-kt \)
(C) ln ([R]0/[R]) vs t — This matches (IV) Straight line passing through origin with positive slope (k). The equation has the form y = mx with slope k.
\( \displaystyle \ln\frac{[R]_0}{[R]}=kt \)
(D) log ([R]0/[R]) vs t — This matches (I) Straight line passing through origin with positive slope (k/2.303). Changing to base-10 logarithms divides the slope by 2.303.
\( \displaystyle \log\frac{[R]_0}{[R]}=\frac{k}{2.303}\,t \)
Shortcut: The pair (A)–(II) appears only in Option (2).