On increasing the pressure three fold, the rate of reaction of `2H_(2)S+O_(2)to"products would incerease"`

Options

A. 3 times
B. 39 times
C. 12 times
D. 27 times

Explanation

To solve the problem of how the rate of reaction changes when the pressure is increased threefold for the reaction \(2H_2S + O_2 \rightarrow \text{products}\), we can follow these steps:

### Step 1: Write the Rate Law
For the reaction \(2H_2S + O_2 \rightarrow \text{products}\), the rate law can be expressed in terms of the concentrations (or partial pressures) of the reactants. The rate law is given by:
\[ \text{Rate} = k [H_2S]^2 [O_2] \]
Where \(k\) is the rate constant.

### Step 2: Convert Concentrations to Partial Pressures
Since we are dealing with gases, we can express the rate in terms of partial pressures instead of concentrations:
\[ \text{Rate} = k (P_{H_2S})^2 (P_{O_2}) \]
Where \(P_{H_2S}\) and \(P_{O_2}\) are the partial pressures of hydrogen sulfide and oxygen, respectively.

### Step 3: Determine the Effect of Increasing Pressure
If we increase the pressure of both \(H_2S\) and \(O_2\) threefold, we can express the new pressures as:
\[ P'_{H_2S} = 3P_{H_2S} \]
\[ P'_{O_2} = 3P_{O_2} \]

### Step 4: Substitute the New Pressures into the Rate Law
Now, we substitute the new pressures into the rate law to find the new rate:
\[ \text{New Rate} = k (P'_{H_2S})^2 (P'_{O_2}) = k (3P_{H_2S})^2 (3P_{O_2}) \]

### Step 5: Simplify the Expression
Now, we simplify the expression:
\[ \text{New Rate} = k (9P_{H_2S}^2) (3P_{O_2}) = 27k (P_{H_2S})^2 (P_{O_2}) \]
Thus, the new rate can be expressed in terms of the original rate:
\[ \text{New Rate} = 27 \times \text{Original Rate} \]

### Conclusion
Therefore, when the pressure is increased threefold, the rate of the reaction increases by a factor of 27.

### Final Answer
The rate of reaction increases to \(27\) times its original rate.

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