Question

The enzyme urease enhances the rate of urea hydrolysis at pH 8.0 and 20C by a factor of 1014 . If a given quantity of urease can completely hydrolyze a given quantity of urea in 5.0 min at 20 C 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? Assume that both reactions take place in sterile systems so that bacteria cannot attack the urea.

Answer 1

Answer:

It will take 950 million years for this amount of urea to be hydrolyzed under the same conditions in the absence of urease.

Explanation:

Given that:

• In the presence of urease:

Urease enhances the rate of hydrolysis by = 10^14

Time taken in hydrolysis is 5 min

• In the absence of urease:

Time taken in hydrolysis will be = 5 min x 10^14

Now, converting minutes into years

 Time   =  (5 min x 10^14) / (60 min/hr x 24 hr/day x 365 days/year)

Time = 9.50 x 10^8 years

Time = 950 X 10^6 years

Time = 950 million years

Answer 2

Answer:

• It would take $5*10^{14}min$ for this amount of of urea to be hydrolysed under the same conditions in the absence of urease.

Explanation:

The enzyme increases the rate of urea hydrolysis by a factor of $10^{14}$

So in absence of enzyme, reaction will take $10^{14}$ times more time for completion.

Therefore time taken,

$\frac{rate catalysed}{rate uncatalysed} = 10^{14}$

$\frac{t uncatalysed}{t catalysed} = 10^{14}\\\\t uncatalysed = 10^{14} * 5min\\\\t uncatalysed = 5*10^{14}min$

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