`Zn+2H^(o+)rarrZn^(2+)+H_(2)`
The half-life periof is independent of the concentration of zinc at constant `pH`. For the constant concentration of `Zn`, the rate becomes `100` times when `pH` is decreased form `3` to `2`. Hence,

Options

A. `(dx)/(dt) = k[Zn]^(0)[H^(o+)]^(2)`
B. `(dx)/(dt) = k[Zn][H^(o+)]^(2)`
C. Rate is not affected if the concentration of zinc is made four times and that of `H^(o+)` ion is halved.
D. (d) Rate becomes four times if the concentration of `H^(o+)` ion is doubled at constant `Zn` concentration.

Explanation

To solve the problem step-by-step, we need to analyze the given reaction and the implications of the changes in pH on the rate of reaction.

### Step 1: Understand the Reaction
The reaction given is:
\[ \text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2 \]

### Step 2: Determine the Rate Law
From the information provided, we know that the half-life period is independent of the concentration of zinc at constant pH. This suggests that the reaction is zero-order with respect to zinc. Therefore, the rate law can be expressed as:
\[ \text{Rate} = k[\text{Zn}]^0[\text{H}^+]^n \]
This simplifies to:
\[ \text{Rate} = k'[\text{H}^+]^n \]
where \( k' = k[\text{Zn}]^0 \).

### Step 3: Analyze the Effect of pH on Rate
The problem states that the rate becomes 100 times greater when the pH decreases from 3 to 2. The pH is related to the concentration of hydrogen ions:
- At pH 3: \([H^+] = 10^{-3} \, \text{M}\)
- At pH 2: \([H^+] = 10^{-2} \, \text{M}\)

Let’s denote the initial rate at pH 3 as \( R_1 \) and the rate at pH 2 as \( R_2 \):
\[ R_1 = k' [10^{-3}]^n \]
\[ R_2 = k' [10^{-2}]^n \]

### Step 4: Relate the Rates
According to the problem:
\[ R_2 = 100 R_1 \]
Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ k' [10^{-2}]^n = 100 \cdot k' [10^{-3}]^n \]

### Step 5: Simplify the Equation
Cancelling \( k' \) from both sides (since it is constant) gives:
\[ [10^{-2}]^n = 100 \cdot [10^{-3}]^n \]
\[ 10^{-2n} = 100 \cdot 10^{-3n} \]
\[ 10^{-2n} = 10^{2} \cdot 10^{-3n} \]
\[ 10^{-2n} = 10^{2 - 3n} \]

### Step 6: Set the Exponents Equal
Since the bases are the same, we can set the exponents equal to each other:
\[ -2n = 2 - 3n \]

### Step 7: Solve for n
Rearranging gives:
\[ 3n - 2n = 2 \]
\[ n = 2 \]

### Step 8: Conclusion
Thus, the order of the reaction with respect to hydrogen ions \( [H^+] \) is 2.

### Final Answer
The correct option is that the order of the reaction with respect to \( [H^+] \) is 2.

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