Question

In the first order reaction t99.99%=xt90%

Answer 1

Explanation:

For t

90%

,a

t

=

100

10a

0

=

10

a

0

t

90%

=

k

1

In

a

0

/10

a−0

=

k

In10

....(1)

For t

99%

, a

t

=

100

1

a−0

t

99%

=

K

1

In

a

0

/100

a

0

=

k

2In10

....(2)

From (1) and (2),

t

99%

=2×t

90%

=>x=2

Answer 2

"It seems this is what you are looking for"

In the first order reaction t99.99%=x(t90%). What will be the value of x?

Answer:

The value of x will be equal to 4.

Explanation:

When the 99.99% reaction is completed:

$t=\frac{2.303}{k} log \frac{A_{o}}{A_{t}}$

$A_{o}$ is an initial concentration of reactants

$A_{t}$ is after time t concentration of reactant

$t_{99.99}=\frac{2.303}{k} log \frac{100}{0.01}$

$t_{99.99}=\frac{2.303}{k} log 10^{4}$                                                         ...........(1)

When 90% of reaction is completed:

$t_{90}=\frac{2.303}{k} log \frac{100}{10}$

$t_{90}=\frac{2.303}{k} log 10$                                                              ............(2)

On dividing eq. (1) by eq.(2)

$\frac{t_{99.99}}{t_{90}}=\frac{log10^{4}}{log10}$

$\frac{t_{99.99}}{t_{90}}=4*\frac{log10}{log10}$

$\frac{t_{99.99}}{t_{90}}=4$

Therefore the value of x is 4.