For a chemical reaction `Y_(2)+2Zto` Product rate controlling step is `Y+1//2ZtoQ`. If the concenntration of Z is doubled the rate of reactin will

Options

A. Remain the same
B. Become four times
C. Become 1.414 times
D. Become double

Answer

To solve the problem, we need to analyze the given chemical reaction and the rate-controlling step. Let's break it down step by step.

### Step 1: Identify the Rate-Controlling Step
The rate-controlling step provided is:
\[ Y + \frac{1}{2}Z \rightarrow Q \]

### Step 2: Write the Rate Expression
The rate of the reaction can be expressed in terms of the concentrations of the reactants. For the rate-controlling step, the rate expression can be written as:
\[ \text{Rate} = k [Y]^a [Z]^b \]
Where \( a \) and \( b \) are the stoichiometric coefficients of \( Y \) and \( Z \) in the rate-controlling step. Here, \( a = 1 \) and \( b = \frac{1}{2} \).

Thus, the rate expression becomes:
\[ \text{Rate} = k [Y]^1 [Z]^{1/2} \]
or simply:
\[ \text{Rate} = k [Y] [Z]^{1/2} \]

### Step 3: Define Initial Conditions
Let’s denote the initial concentration of \( Z \) as \( [Z] \). Therefore, the initial rate \( r_1 \) can be expressed as:
\[ r_1 = k [Y] [Z]^{1/2} \]

### Step 4: Change the Concentration of \( Z \)
According to the problem, the concentration of \( Z \) is doubled. Thus, the new concentration of \( Z \) will be \( 2[Z] \).

### Step 5: Calculate the New Rate
Now, we can calculate the new rate \( r_2 \) with the doubled concentration of \( Z \):
\[ r_2 = k [Y] (2[Z])^{1/2} \]

### Step 6: Simplify the New Rate Expression
Now, simplify the expression for \( r_2 \):
\[ r_2 = k [Y] (2^{1/2} [Z]^{1/2}) \]
\[ r_2 = k [Y] \sqrt{2} [Z]^{1/2} \]
\[ r_2 = \sqrt{2} \cdot k [Y] [Z]^{1/2} \]

### Step 7: Relate the New Rate to the Initial Rate
Since \( r_1 = k [Y] [Z]^{1/2} \), we can substitute this into the equation for \( r_2 \):
\[ r_2 = \sqrt{2} \cdot r_1 \]

### Step 8: Conclusion
Thus, when the concentration of \( Z \) is doubled, the rate of reaction becomes:
\[ r_2 = \sqrt{2} \cdot r_1 \approx 1.414 \cdot r_1 \]

Therefore, the rate of the reaction will increase by a factor of approximately 1.414 times.

### Final Answer
When the concentration of \( Z \) is doubled, the rate of reaction will become approximately 1.414 times the original rate.

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