At a certain temperature, the half-life periods for the catalytic decomposition of NH3 were found to be as follows :- 0.88 What will be the pressure when the half-life period is 1.5 hours.

Question

At a certain temperature, the half-life periods for the catalytic decomposition of NH3 were found to be as follows :- 0.88 What will be the pressure when the half-life period is 1.5 hours. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(( t 1 / 2)1/(t1 / 2)2)=((P2/P1))n-1 from given data (8.52/1.96)=((100/50)) n -1 2=2 n -1 n -1=1 n =2 Again (3.52/1.5)=(( P /50))2-1 ( P /50)=(3.52/1.5) P =50 × 2.34 P =117

Q. At a certain temperature, the half-life periods for the catalytic decomposition of $N H_{3}$ were found to be as follows :-
$0.88$

What will be the pressure when the half-life period is $1.5$ hours.

$\frac{( t _{1/2} ) _{1}}{( t _{1/2} ) _{2}} = (\frac{P _{2}}{P _{1}})^{n - 1}$ from given data
$\frac{8.52}{1.96} = (\frac{100}{50})^{n - 1}$
$2 = 2^{n - 1}$
$n - 1 = 1$
$n = 2$
Again $\frac{3.52}{1.5} = (\frac{P}{50})^{2 - 1} \frac{P}{50} = \frac{3.52}{1.5}$
$P = 50 \times 2.34$
$P = 117$

Q. At a certain temperature, the half-life periods for the catalytic decomposition of NH3​ were found to be as follows :- 0.88 What will be the pressure when the half-life period is 1.5 hours.

(t1/2​)2​(t1/2​)1​​=(P1​P2​​)n−1 from given data 1.968.52​=(50100​)n−1 2=2n−1 n−1=1 n=2 Again 1.53.52​=(50P​)2−150P​=1.53.52​ P=50×2.34 P=117

Q. At a certain temperature, the half-life periods for the catalytic decomposition of $N H_{3}$ were found to be as follows :-
$0.88$

What will be the pressure when the half-life period is $1.5$ hours.

$\frac{( t _{1/2} ) _{1}}{( t _{1/2} ) _{2}} = (\frac{P _{2}}{P _{1}})^{n - 1}$ from given data
$\frac{8.52}{1.96} = (\frac{100}{50})^{n - 1}$
$2 = 2^{n - 1}$
$n - 1 = 1$
$n = 2$
Again $\frac{3.52}{1.5} = (\frac{P}{50})^{2 - 1} \frac{P}{50} = \frac{3.52}{1.5}$
$P = 50 \times 2.34$
$P = 117$