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Standard XII

Chemistry

Preparations and Reactions of Sodium Oxide

The time requ...

Question

The time required for $15$ % decomposition at an initial conc. of $0.1 M$ in a first order reaction $A \to P r o d u c t s$ , at $80^{\circ} C$ is $25 m i n$ , then the time required for decomposition by $0.06 M$ at an initial conc. of $0.4 M$ at $353 K$ is:

5 m i n

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0.25 m i n

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25 \times l n 2 m i n

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25 m i n

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Solution

The correct option is A $5 m i n$
If $A_{0}$ decomposes 15% then $A = 0.85 A_{0}$
Thus, $\frac{1}{t} l n (\frac{A_{0}}{A}) = k \Rightarrow k = 8.126 \times 10^{- 3} m i n^{- 1}$
now using $t = \frac{1}{k} l n (\frac{A_{0}}{A}) :$
$t = \frac{1}{8.126 \times 10^{- 3}} \times l n (\frac{0.4}{0.06}) = 5$ min

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The time required for 15% decomposition at an initial conc. of 0.1 M in a first order reaction A→Products, at 80∘C is 25 min, then the time required for decomposition by 0.06 M at an initial conc. of 0.4 M at 353 K is:

The correct option is A 5 minIf A0 decomposes 15% then A=0.85A0Thus,1tln(A0A)=k⇒k=8.126×10−3min−1now using t=1kln(A0A):t=18.126×10−3×ln(0.40.06)=5 min

The time required for $15$ % decomposition at an initial conc. of $0.1 M$ in a first order reaction $A \to P r o d u c t s$ , at $80^{\circ} C$ is $25 m i n$ , then the time required for decomposition by $0.06 M$ at an initial conc. of $0.4 M$ at $353 K$ is: