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When inversion of sucrose is studied at ...

When inversion of sucrose is studied at pH = 5, the half-life period is always found to be 500 minutes irrespective of any initial concentration but when it is studied at pH = 6, the half-life period is found to be 50 minutes. Derive the rate law expression for the inversion of sucrose.

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To derive the rate law expression for the inversion of sucrose based on the provided information, we can follow these steps: ### Step 1: Understand the Reaction The inversion of sucrose (C12H22O11) in the presence of water results in the formation of glucose (C6H12O6) and fructose (C6H12O6). The reaction can be represented as: \[ \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6 \] ### Step 2: Identify the Order of Reaction The problem states that at pH = 5, the half-life (t₁/₂) is 500 minutes, and at pH = 6, the half-life is 50 minutes. Since the half-life is independent of the initial concentration at pH = 5, we can conclude that the reaction is first-order with respect to sucrose. ...
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