Match order of the reaction (in List-I) with the corresponding rate constant (in List II).
`{:("List I (order)","List II (rate constant)"),((A)"Zero",(p)K=1/(2t)[1/((a-x)^2)-1/a^2]),((B)"First",(q)K=1/(t)[1/((a-x))-1/a]),((C )"Second",(r)k=x/t),((D)"Third",(s)k=1/t"log"_e(a/(a-x))):}`

Explanation

To solve the problem of matching the order of the reaction with the corresponding rate constant, we will analyze each order of reaction and derive the respective rate constant expressions.

### Step-by-Step Solution:

1. **Zero Order Reaction**:
- For a zero-order reaction, the rate law is given by:
\[ [A] = [A_0] - kt \]
- Rearranging gives:
\[ k = \frac{x}{t} \]
- Here, \(x\) is the amount of reactant that has reacted, and \(t\) is the time.
- Thus, for zero order, we match:
- **A → r** (k = x/t)

2. **First Order Reaction**:
- For a first-order reaction, the integrated rate law is:
\[ \ln\left(\frac{[A_0]}{[A]}\right) = kt \]
- Rearranging gives:
\[ k = \frac{1}{t} \ln\left(\frac{[A_0]}{[A]}\right) = \frac{1}{t} \ln\left(\frac{a}{a-x}\right) \]
- Thus, for first order, we match:
- **B → s** (k = 1/t * log_e(a/(a-x)))

3. **Second Order Reaction**:
- For a second-order reaction, the integrated rate law is:
\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]
- Rearranging gives:
\[ k = \frac{1}{t}\left(\frac{1}{[A]} - \frac{1}{[A_0]}\right) = \frac{1}{t}\left(\frac{1}{a-x} - \frac{1}{a}\right) \]
- Thus, for second order, we match:
- **C → q** (k = 1/t * [1/(a-x) - 1/a])

4. **Third Order Reaction**:
- For a third-order reaction, the integrated rate law is:
\[ \frac{1}{[A]^2} - \frac{1}{[A_0]^2} = kt \]
- Rearranging gives:
\[ k = \frac{1}{t}\left(\frac{1}{(a-x)^2} - \frac{1}{a^2}\right) \]
- Thus, for third order, we match:
- **D → p** (k = 1/(2t) * [1/(a-x)^2 - 1/a^2])

### Summary of Matches:
- A → r (Zero order)
- B → s (First order)
- C → q (Second order)
- D → p (Third order)

### Final Answer:
- The final matching is:
- A - r
- B - s
- C - q
- D - p