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The order of a reaction with rate equal ...

The order of a reaction with rate equal to `KC_(A)^(1//2)C_B^(-1//2)` is

A

2

B

Zero

C

`-1/2`

D

`3/2`

Text Solution

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The correct Answer is:
To determine the order of the reaction given the rate expression \( \text{Rate} = k C_A^{1/2} C_B^{-1/2} \), we will follow these steps: ### Step 1: Identify the Rate Law Expression The rate law expression is given as: \[ \text{Rate} = k C_A^{1/2} C_B^{-1/2} \] Here, \( C_A \) and \( C_B \) are the concentrations of reactants A and B, respectively, and \( k \) is the rate constant. ### Step 2: Determine the Exponents In the rate expression: - The exponent of \( C_A \) is \( \frac{1}{2} \). - The exponent of \( C_B \) is \( -\frac{1}{2} \). ### Step 3: Calculate the Overall Order of the Reaction The overall order of the reaction is the sum of the exponents of the concentrations in the rate law. Therefore, we calculate: \[ \text{Order} = \left(\frac{1}{2}\right) + \left(-\frac{1}{2}\right) \] ### Step 4: Simplify the Expression Now, we simplify the expression: \[ \text{Order} = \frac{1}{2} - \frac{1}{2} = 0 \] ### Conclusion The overall order of the reaction is 0. Thus, the order of the reaction is **0**. ---
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