Match List-I with List-II
|
List-I Reaction property |
List-II Formula |
|
(A) Half-life of decomposition of gaseous ammonia on hot platinum surface |
(I) $\frac{ln(\frac{a_0}{a_t})}{t}$ |
|
(B) Instantaneous rate |
(II) $[R]_0/2k$ |
|
(C) Rate constant of inversion of cane sugar |
(III)$\frac{dx}{dt}$ |
|
(D) Half-life of acid hydrolysis of methyl acetate |
(IV) $(ln2)/k$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
|
List-I Reaction property |
List-II Formula |
|
(A) Half-life of decomposition of gaseous ammonia on hot platinum surface |
(II) $[R]_0/2k$ |
|
(B) Instantaneous rate |
(III)$\frac{dx}{dt}$ |
|
(C) Rate constant of inversion of cane sugar |
(I) $\frac{ln(\frac{a_0}{a_t})}{t}$ |
|
(D) Half-life of acid hydrolysis of methyl acetate |
(IV) $(ln2)/k$ |
(A) Half-life of decomposition of gaseous ammonia on hot platinum surface:
The decomposition of $NH_3$ on a platinum surface at high pressure is a classic example of a zero-order reaction. The rate is independent of the concentration because the catalyst surface becomes saturated.
- Zero-order half-life formula: $t_{1/2} = \frac{[R]_0}{2k}$
- Matches with: (II)
(B) Instantaneous rate:
In calculus-based kinetics, the instantaneous rate is the rate of reaction at a specific moment, represented by the derivative of concentration with respect to time.
- Formula: $\frac{dx}{dt}$
- Matches with: (III)
(C) Rate constant of inversion of cane sugar:
The inversion of cane sugar (sucrose) is a pseudo-first-order reaction. The rate constant ($k$) for a first-order reaction is derived from the integrated rate equation: $\ln\left(\frac{a_0}{a_t}\right) = kt$.
- Formula: $k = \frac{\ln(a_0/a_t)}{t}$
- Matches with: (I)
(D) Half-life of acid hydrolysis of methyl acetate:
Acid hydrolysis of esters like methyl acetate is also a pseudo-first-order reaction because water is present in such large excess that its concentration remains constant.
- First-order half-life formula: $t_{1/2} = \frac{\ln 2}{k}$ (or $\frac{0.693}{k}$)
- Matches with: (IV)