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The hydrolysis of sucrose to glucose and fructose is catalyzed by the enzyme sucrase. The hydrolysis is first order in sucrose with a half-life of 4.8 103 s at 20°C. What fraction of sucrose remains after 2.0 hours? Please see Half Life of First Order Reactions for assistance. WebAssign will check your answer for the correct number of significant figures. .646 Incorrect: Your answer is incorrect. How long is required to hydrolyze 94% of the sucrose? WebAssign will check your answer for the correct number of significant figures. 54.1 Incorrect: Your answer is incorrect. hr

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Answer:

  • Question 1: 0.35

  • Question 2: 5.4  hour

Explanation:

The rate law of a first-order reaction is:

      [tex]rate=-\dfrac{d[A]}{dt}=k[A][/tex]

Which, after integration, becomes:

         [tex]\dfrac{[A]}{[A]_0}=e^{-kt}[/tex]

The half-life means [A]/[A]₀ = 0.5. Then:

          [tex]0.5=e^{-kt_{1/2}}[/tex]

From which:

         [tex]k=\dfrac{\ln 2}{t_{1/2}}[/tex]

Since, the half-life is a constant, you can write the final concentration in terms of the number of half-lives elapsed, n:

           [tex]\dfrac{A}{[A]_0}=\bigg(\dfrac{1}{2}\bigg)^n[/tex]

Question 1:

Since the half-life is 4.8 × 10³s and you want the fraction after 2.0 hours:

  • n = 2.0hour × 3,600s/hour / 4,800s = 1.5

  • [A]/[A]₀ = 0.354 = 0.35 ← answer to the first question.

Question 2:

  • Hidrolize 94% ⇒ [A]/[A]₀ = 1 - 0.94 = 0.06

  • n = log (0.06) / log (0.5)

  • n = 4.05889

  • time = n × 4,800 s = 4.05889 × 4,800s =  19,482.7s

  • time = 19,482.7 s × 1hour/3600s = 5.4hour ← answer to the second question.

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