Question

For first order reaction, t93.75% = x * 75%, the value
of x will be
(1) 10
(2) 6
(3) 3
(4) 2.

please give a written explanation , please​

Answer 1

Answer:

For the first-order reaction, $t_{93.75} = x \times t_{75}$ the value of x is 2

Therefore, option d) 2 is the correct answer.

Explanation:

Given,

For the first-order reaction;

Total concentration at time zero, A₀= 100 %

Concentration after completion 93.75% at a time ( t ) = ( 100 - 93.75 )%

Concentration after completion 75% at a time ( t ) = ( 100 - 75 ) %

For first order reaction,$t_{93.75} = x \times t_{75}$ the value of x =?

As we know,

• The half-life for the first-order reaction can be calculated by the equation given below:
• $t = \frac{2.303}{k} log[\frac{A_{0} }{A }]$

For completion of 93.75%:

• $t_{93.75} = \frac{2.303}{k} log[\frac{100 }{100 - 93.75}]$
• $t_{93.75} = \frac{2.303}{k} log[\frac{100 }{6.25}]$    ------------------------ equation (1)

For completion of 75%:

• $t_{75} = \frac{2.303}{k} log[\frac{100 }{100 - 75}]$
• $t_{75} = \frac{2.303}{k} log[\frac{100 }{25}]$   ------------------------- equation (2)

Now, after dividing equation 1 by 2, we get:

• $\frac{t_{93.75} = \frac{2.303}{k} log[\frac{100 }{6.25}]}{t_{93.75} = \frac{2.303}{k} log[\frac{100 }{6.25}]}$

• $\frac{t_{93.75} }{t_{75} } = \frac{log16}{log4}$

• $\frac{t_{93.75} }{t_{75} } = \frac{log2^{4} }{log2^{2} }$
• $\frac{t_{93.75} }{t_{75} } = \frac{4log2}{2log2}$
• $\frac{t_{93.75} }{t_{75} } = \frac{4}{2}$
• $\frac{t_{93.75} }{t_{75} } = 2$

• $t_{93.75} = 2 \times t_{75}$

Hence, the value of x = 2

Answer 2

Answer:

HOPE THE ANSWER HELPS YOU!!