$C_{12} H_{22} O_{11} + H_{2} O \to C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}$
(excess) (gulcose) (fructose)
Rate law of above equation is expressed as:

r = k [C_{12} H_{22} O_{11}] [H_{2} O]

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r = k [C_{12} H_{22} O_{11}]

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r = k [H_{2} O]

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r = k [C_{12} H_{22} O_{11}] {[H_{2} O]}_{2}^{2}

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Solution

The correct option is A $r = k [C_{12} H_{22} O_{11}]$
In the reaction, $C_{12} H_{22} O_{11} + H_{2} O \to C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}$

Here, $H_{2} O$ is present in excess.

Hence, it will not appear in the rate law expression. Thus, it is an example of a pseudo-first-order reaction.

The rate law expression for the reaction is, $r = k [C_{12} H_{22} O_{11}]$ .

Option B is correct.

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Similar questions

C_{12} H_{22} O_{11} Sucrose + H_{2} O Enzyme A - ----- \to C_{6} H_{12} O_{6} Glucose + 2 C_{6} H_{12} O_{6} Fructose

C_{6} H_{12} O_{6} Glucose Enzyme B - ----- \to 2 C_{2} H_{5} O H + 2 C O_{2}

C_{12} H_{22} O_{11} + H_{2} O i n v e r t a s e - ---- \to . . . . . . . . + . . . . . . . .

C_{12} H_{22} O_{11} C a n e s u g a r + H_{2} O H^{+} - - \to C_{6} H_{12} O_{6} G l u c o s e + C_{6} H_{12} O_{6} F r u c t o s e

H_{2} O

C_{12} H_{22} O_{11} + H_{2} O \to C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}

^{o}

(K_{f} (H_{2} O) = 1.86 K k g m o l^{1})

(C_{12} H_{22} O_{11})

C_{12} H_{22} O_{11} d-Sucrose + H_{2} O H^{+} - - \to C_{6} H_{12} O_{6} d -Glucose + C_{6} H_{12} O_{6} l -Fructose

r_{0}

r_{t}

r_{\infty}

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