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Standard X

Chemistry

Factors Affecting the Rate of a Chemical Reaction

In order to e...

Question

In order to economize on the oxygen supply in space ships, it has been suggested that the oxygen in exhaled $C O_{2}$ be converted to water by a reduction with hydrogen. The $C O_{2}$ output per astronaut has been estimated as $1.00 k g$ per $24 h r$ day.
An experimental catalytic converter reduces $C O_{2}$ at a rate of $600 m L$ (STP) per minute.
What fraction of the time would such a converter have to operate in order to keep up with the $C O_{2}$ output of one astronaut.

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Solution

Given that,
$C O_{2}$ output per astronaut= $1 k g$ per $24$ hours
Now,
catalytic converter reduces $C O_{2}$ at rate of $600 m L$
$(S T P)$ per minute
At $(S T P)$ , $T = 273 K$ , $P = 1$ atm
$R = 0.0821$ $L$ atm $K^{- 1} m o l^{- 1}$
Given, $V = 600 m L = 0.6 L$
By Ideal gas $e q^{n}$ ,
$P V = n R T$
$\Rightarrow n = \frac{P V}{R T}$
$\Rightarrow n = \frac{1 \times 0.6}{0.0821 \times 273} = 0.028$ moles /minute
No. of moles of $C O_{2}$ output by astronaut
$= \frac{1000 g}{46 g} = 21.74$ per $24$ hours
$(∵ N o . o f m o l e s = w e i g h t / M o l e c u l a r$ $w e i g h t$ & Molar mass of $C O_{2} = 44 g)$
So, time required= $\frac{21.74}{0.024}$
= $905.8$ minutes per day
So, time of operation of catalytic convertor in $24$ hours= $15.1$ hours

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