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The inversion of cane sugar proceeds with half- life of 500 minutes at pH=5 for any concentration of sugar. However, if pH=6, the half-life changes to 50 minutes. Derive the rate law for inversion of cane-sugar. |
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Answer» Solution :At `pH=5`, the half-life is `500` for all CONCENTRATIONS of sugar, i.e. `t_(1//2) PROP ["sugar"]^(0)`. Thus, the REACTION is I order with respect to sugar, Now rate `=K["sugar"]^(1)[H^(+)]^(m)` Also, for `[H^(+)]t_(1//2) prop [H^(+)]^(1-m)` `500 prop [10^(-5)]^(1-m)` `500 prop [10^(-6)]^(1-m)` `10=(10)^(1-m)` `1-m=1` `m=0` Threfore, rate `-K["sugar"]^(1)[H^(+)]^(0)` |
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