The enzyme urease enhaces the rate of urea hydrolysis at pH8.0 and 20 degree of centigrade by a factor of 1014. If a given quantity of urea in 5 min at pH 8.0 and 20 degree of centigrade, how long would it take for this amount of urea to be hydrolyzed under the same conditions in the absence of urease

In the absence of urease this will take 950 * 10^6 years to  completely hydrolyze

Explanation:

The enzyme urease enhances the rate of urea hydrolysis at pH 8.0 and 20 degrees by a factor of 10^14 . If a given quantity of urease can completely hydrolyze a given quantity of urea in 5 min at 20 degrees and pH 8.0, how long would it take for this amount of urea to be hydrolyzed under the same conditions in the absence of urease?

Step 1: data given

The enzyme urease enhaces the rate of urea hydrolysis at pH = 8.0 of centigrade by a factor of 10^14

Temperature = 20.0 °C

Step 2: Calculate the time it takes to completely hydrolyze without urease

This means when urease is available it takes 5 minutes to completely hydrolyze

When urease is not available it will take 5 *10^14 minutes

5 * 10^14 min = 5 * 10^14 minutes

Since 1 hour = 60 minutes

1 day = 24 hours = 24 *60 minutes = 1440 minutes

It would take :  5 * 10^14 minutes / 1440 minutes = 347222222222.2 days

Since 1 year = 365 days

This will take: 347222222222.2 / 365.25 = 950642634.4 years

This 950 * 10^6 years

In the absence of urease this will take 950 * 10^6 years

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.

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