10 gm mixture of two gases A 2 ( Mol . wt . = 20 ) and B 2 ( Mol.wt. = 30 ) , which decompose by first order kinetics, was taken in a vessel. The half life for decomposition of A 2 and B 2 are

Question

10 gm mixture of two gases A 2 ​ ( Mol . wt . = 20 ) and B 2 ​ ( Mol.wt. = 30 ) , which decompose by first order kinetics, was taken in a vessel. The half life for decomposition of A 2 ​ and B 2 ​ are 2 and 3 hours respectively. After 6 hours, weight of the gases A 2 ​ and B 2 ​ is found to be 2 gm . Find the weight (in mg ) of remaining A 2 ​ & B 2 ​ gases after 12 hours.

TestHub · Accepted Answer

Initial setup: Let the initial masses of gases A₂ and B₂ be $x$ and $y$ grams, respectively, where $x + y = 10 g m$ .

Mass after 6 hours: Since the half-lives are 2 hours for A₂ and 3 hours for B₂, after 6 hours A₂ completes 3 half-lives and B₂ completes 2 half-lives:

$\frac{x}{2 ^{3}} + \frac{y}{2 ^{2}} = 2 g m ⟹ \frac{x}{8} + \frac{y}{4} = 2 ⟹ x + 2 y = 16$

Solving for initial masses: Subtracting the first equation from this result gives $y = 6 g m$ and $x = 4 g m$ .

Remaining mass after 12 hours: After 12 hours, A₂ completes 6 half-lives and B₂ completes 4 half-lives:

$R e mainin g M a ss = \frac{4}{2 ^{6}} + \frac{6}{2 ^{4}} = \frac{4}{64} + \frac{6}{16} = \frac{1}{16} + \frac{6}{16} = \frac{7}{16} g m$

Final Answer: Converting to milligrams gives $\frac{7}{16} \times 1000 = 437.50 m g$ .

Chemistry - Chemical Kinetics Question with Solution | TestHub

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