Question

A first order rxn. Has a rate constant of 2.303×10^-3 per second. The time required for 40 g of this reached to reduce 10 g will be:- A. 2.303 sec
B. 301 sec
C. 2000 sec
D. 602 sec ( plzzz answer as soon as Possible) ( in great need of the answer)

Answer 1

Answer:

Option D is the right option.

Explanation:

Refer attachment for explanation!!

Answer 2

Answer:

The time required, t, for the reaction is $602sec$.

Therefore, option d) $602sec$ is correct.

Explanation:

Given,

The rate constant for a first-order reaction, k = $2.303 \times 10^{-3} s^{-1}$

The reaction's initial concentration, [A] = $40g$

The final concentration after reduction, [A°] = $10g$

The time required for the reaction, t =?

As we know,

• Integrated rate law expression for the first-order reaction is:
• $k = \frac{2.303}{t} log\frac{[A]}{[A\textdegree]}$
• $t = \frac{2.303}{k} log\frac{[A]}{[A\textdegree]}$

After putting the given values in the equation, we get:

• $t = \frac{2.303}{2.303\times 10^{-3} } log\frac{40}{10}$
• $t = 10^{3} log 4$
• t = $10^{3}\times 0.602$                ( $log 4 = 0.602$ )
• t = $602 sec$

Hence, the time required for the reaction, t = $602 sec$.