For the reaction: `X(g) rarr Y(g)+Z(g)`, the following data were obtained at `30^(@)C`:
`{:("Experiment",[X](mol L^(-1)),"Rate" (mol L^(-1) hr^(-1))),(I,0.17,0.05),(II,0.34,0.10),(III,0.68,0.20):}`
The equilibrium constant for the reaction is `0.50`. Assuming that the reaction proceeds by one-step mechanism. Find the rate constant of reverse reaction?

Options

A. `0.294 hr^(-1)`
B. `0.588 hr^(-1)`
C. `0.123 hr^(-1)`
D. `0.117 hr^(-1)`

Explanation

To solve the problem step by step, we need to find the rate constant of the reverse reaction for the given reaction \( X(g) \rightarrow Y(g) + Z(g) \).

### Step 1: Determine the order of the reaction
From the given data, we can use the rates and concentrations from two experiments to find the order of the reaction.

Using experiments I and II:
\[ \frac{\text{Rate}_I}{\text{Rate}_{II}} = \frac{0.05}{0.10} = \frac{1}{2} \]
\[ \frac{[X]_I^a}{[X]_{II}^a} = \frac{0.17^a}{0.34^a} = \left(\frac{1}{2}\right) \]

This simplifies to:
\[ \frac{1}{2} = \left(\frac{1}{2}\right)^a \]
From this, we can conclude that \( a = 1 \). Thus, the order of the reaction is 1.

### Step 2: Write the rate law expression
Since the order of the reaction is 1, we can express the rate law as:
\[ \text{Rate} = k_f [X] \]
where \( k_f \) is the rate constant for the forward reaction.

### Step 3: Calculate the rate constant for the forward reaction
Using the data from experiment I:
\[ \text{Rate} = 0.05 \, \text{mol L}^{-1} \text{hr}^{-1}, \quad [X] = 0.17 \, \text{mol L}^{-1} \]
Substituting these values into the rate law:
\[ 0.05 = k_f \times 0.17 \]
Solving for \( k_f \):
\[ k_f = \frac{0.05}{0.17} \approx 0.294 \, \text{hr}^{-1} \]

### Step 4: Use the equilibrium constant to find the rate constant for the reverse reaction
The equilibrium constant \( K \) is given by:
\[ K = \frac{k_f}{k_b} \]
where \( k_b \) is the rate constant for the reverse reaction. We know that \( K = 0.50 \) and \( k_f \approx 0.294 \, \text{hr}^{-1} \).

Rearranging the equation to solve for \( k_b \):
\[ k_b = \frac{k_f}{K} = \frac{0.294}{0.50} \approx 0.588 \, \text{hr}^{-1} \]

### Final Answer
The rate constant of the reverse reaction \( k_b \) is approximately \( 0.588 \, \text{hr}^{-1} \).

---