Question

The half life period of pyruvic acid in the presence of an amino transferase enzyme (which converts
in to alanine) was found to be 221 sec. How long will it take for the concentration of pyruvic acid to
fall to 1/64 of its initial value in this first order reaction?
1) 1200 sec
2) 2440 sec
3) 663 se
4) 1326 sec​

Answer 1

Answer:

2440 sec is the correct answer

Answer 2

Answer:

The time taken, t, for the concentration of pyruvic acid to fall $\frac{1}{64}th$ of its initial value measured is $1326sec$.

Therefore, option 4) $1326sec$ is correct.

Explanation:

Given,

The half-life period of pyruvic acid, $t_{1/2}$ = $221sec$

Let the initial concentration of pyruvic acid, $A_{0}$ = A

Therefore, as given,

The final concentration of pyruvic acid, $A_{t}$ = $\frac{A}{64}$

The time taken, t, for the concentration of pyruvic acid to fall $\frac{1}{64}th$ of its initial value =?

As we know,

• For the first-order reaction;
• t = $\frac{2.303}{k} log \frac{A_{0} }{A_{t} }$   -------equation (1)

Here, k = rate constant

Firstly, we have to find out the value of k.

• $t_{1/2}$ = $\frac{0.693}{k}$
• $221$ = $\frac{0.693}{k}$
• k = $\frac{0.693}{221}$ = $0.003135 sec^{-1}$

After putting the values in equation (1), we get:

• t = $\frac{2.303}{0.003135} log \frac{A }{\frac{A}{64} }$
• t = $\frac{2.303}{0.003135} log \frac{64A}{A} }$
• t = $734.6 log 64$
• t = $734.6 \times 1.806$     ( $log 64 =1.806$ )
• t = $1326.6 sec$ ≈ $1326sec$

Hence, the time taken, t, for the concentration of pyruvic acid to fall $\frac{1}{64}th$ of its initial value = $1326sec$.