Match the LIST-I with LIST-II
\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline A. \ ^{236}_{92} U \rightarrow ^{94}_{38} Sr + ^{140}_{54} Xe + 2n & \text{I. Chemical Reaction} \\ \hline B. \ 2H_2 + O_2 \rightarrow 2H_2O & \text{II. Fusion with +ve Q value} \\ \hline C. \ ^3_1 H + ^2_1 H \rightarrow ^4_2 He + n & \text{III. Fission} \\ \hline D. \ ^1_1 H + ^3_1 H \rightarrow ^4_2 H + \gamma & \text{IV. Fusion with -ve Q value} \\ \hline \end{array} \]
Choose the correct answer from the options given below:
Let's examine the types of reactions listed in LIST-I and match them with their corresponding descriptions in LIST-II.
Thus, the correct matching from LIST-I with LIST-II is as follows:
Therefore, the correct answer is:
A-III, B-I, C-II, D-IV
Therefore, the correct matches are:
A - III
B - I
C - II
D - IV
For a given reaction \( R \rightarrow P \), \( t_{1/2} \) is related to \([A_0]\) as given in the table. Given: \( \log 2 = 0.30 \). Which of the following is true?
| \([A]\) (mol/L) | \(t_{1/2}\) (min) |
|---|---|
| 0.100 | 200 |
| 0.025 | 100 |
A. The order of the reaction is \( \frac{1}{2} \).
B. If \( [A_0] \) is 1 M, then \( t_{1/2} \) is \( 200/\sqrt{10} \) min.
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. \( t_{1/2} \) is 800 min for \( [A_0] = 1.6 \) M.