Question

The initial concentration of cane sugar in the presence of an acid was reduced from 0.2 to 0.1M in 5 hr and to 0.05 in 10hours. value of K is(in hr) ​

Answer 1

Answer:

k = 0.138 hr⁻¹.

Explanation:

It took 5 hours to fall from 0.2 M concentration to 0.1 M concentration (1/2).

It took 10 hours to fall from 0.2 M to 0.05 M (1/4).

Clearly this is first order reaction with half life 5 hours.

And we know that $k = \frac{ln 2}{t_1_/_2}$

Hence k = ln2/5

k = 0.693/5

k = 0.138 hr⁻¹.

Answer 2

Answer:

The value of rate constant (k) is equal to 0.1386 hr⁻¹.

Explanation:

From the unit of rate constant (k) i.e. hr⁻¹ , we can assume that it could be a first order reaction.

Case-I: The initial concentration of cane sugar, $[A_{o}]=0.2M$

           Concentration at time t, $[A_{t}]=0.1M$

            Time, $t=5hr$

           For first order reaction: $ln \frac{[A_{o}]}{[A_{t}]} =kt$

          Therefore, $k=\frac{1}{5} ln\frac{0.2}{0.1}$

          ∴      $k=0.1386hr^{-1}$

Case-II : The initial concentration $[A_{o}]=0.1M$  and $[A_{t}]=0.05M$

               Time, $t=5hr$

               Now,   $k=\frac{1}{5} ln\frac{0.1}{0.05}$

               $k=0.1386hr^{-1}$

So, the value of rate constant is same for both cases. It is concluded that the reaction is first order reaction.

Therefore, the value of k is 0.1386hr⁻¹