The rate of the hydrolysis of sucrose to glucose is quite slow in the absence of

Question

The rate of the hydrolysis of sucrose to glucose is quite slow in the absence ofa catalyst.

Sucrose + water glucose + fructose

a. If the initial concentration of sucrose is 0.050M, it takes 440 years for theinitial concentration of the sucrose to decrease by half to 0.025M. What is therate of disappearance, (M-s-1) of sucrose in the absence of a catalyst?

b. If a catalyst is present, the hydrolysis of sucrose is much more rapid. If theinitial concentration of sucrose is 0.050M, it takes 6.9x10-5 s for theconcentration of sucrose to decrease by half to 0.025M. What is the rate of the disappearance of sucrose in the presence of a catalyst?

c. Determine the velocity of the uncatalyzed reaction when the concentration ofsucrose is 0.050M. The rate law is: rate = k [sucrose]. The rate constant for theuncatalyzed reaction is 5.0 x 10-11 s-1.

d. Determine the velocity of the catalyzed reaction when the concentration ofsucrose is 0.050M. The rate law is: rate = k [sucrose]. The rate constant for thecatalyzed reaction is 1.0 x 104 s-1.

e. The enzyme catalyzed reaction has a KM of 0.135 mM and a Vmax of 65?molmin-1. What is the reaction velocity when the concentration of sucrose is 1.0 mM?

Explanation / Answer

a)

Rate of disappearance = - Change in concentration of sucrose / time in seconds

Change in concentration = Final concentration - Initial concentration

Change in concentration = 0.025 - 0.050 = -0.025 M

time is given in yrs. Let's convert it to seconds

440 yrs * (365 days / 1 yr) * ( 24 hrs / 1 day) * ( 60minutes / 1 hr) * ( 60 seconds/ 1 minute) = 1.39x10^10 seconds

Rate of disappearance = - ( -0.025 M) / 1.39 x 10^10 s = 1.8 x 10^-12 M/s

Rate of disappearance = 1.8 x 10^-12 M/s

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b)

Change in concentration is 0.025 M

Time = 6.9 x 10^-5 s

Using the same formula in part a, we get

Rate of disappearance = - ( -0.025 M) / 6.9 x 10^-5 s

Rate of disappearance = 362 M/s

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c)

The rate law is given as

Rate = k [ Sucrose]

Concentration of sucrose is given as 0.050 M

k = 5.0 x 10^-11 s^-1

Rate = 5.0 x 10^-11 s^-1 * 0.050 M

Rate = 2.5 x 10^-12 M/s

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d)

Concentration of sucrose = 0.05 M

k = 1.0 x 10^4 s^-1

Rate = k [ Sucrose]

Rate = 1.0 x10^4 s^-1 * 0.05 M

Rate = 500 M/s

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e)

We can use Michaelis–Menten equation to solve this.

The equation is written as

Vo = Vmax * [S] / ( Km + [S]) where,

Vo is initial velocity

Km is Michaelis constant

Vmax is maximum velocity

[S] is substrate concentration

From the given data, we have

Vmax = 65 umol/min ( Note : Please check if this value is correct. there is some formatting error in the question)

Km = 0.135 mM

[S] = 1.0 mM

We have to find reaction velocity Vo

Substituting these values in Michaelis–Menten equation

Vo = 65 umol/min * 1.0 mM / ( 0.135 mM + 1.0 mM)

Vo = 65 umol/min * 1 mM/ 1.135 mM

Vo = 57.3 umol/min

1 umol = 10^-6 mol.............Using this conversion factor we get

Vo = 5.73 x 10^-6 mol/min

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